What do continuous maps between $X$ and $Y$ tell us about $X$ and $Y$?Meaning of “a mapping preserves...

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What do continuous maps between $X$ and $Y$ tell us about $X$ and $Y$?


Meaning of “a mapping preserves structures/properties”Which (endo)functors of the category of finite-dimensional real vector spaces induce continuous maps between Hom-sets?Axiomatizing topology through continuous mapsRelations between Theories and CategoriesAnalogy between quotient groups and quotient topologyMonoid and group objects in slice categoriesWhy don't we take clopen maps as morphisms of the category of topological spaces?$p$-complete abelian groups and $p$-continuous homomorphismsCan we rederive the axioms of topology from the structure of the category $textbf{Top}$?Formal definition of “equivalence”between two formalizations of a theory?













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$begingroup$


Category theory, roughly, is about the relation between structures of a certain kind, rather than their internal structure.



In the category of groups, I can see how this is the case: the existence of a particular group homomorphism between $G$ and $H$ really tells you alot about $G$ and $H$ as structures.



But I’ve never thought of continuous maps between topological spaces $X$ and $Y$ as “telling us something about $X$ and $Y$”. Admittedly, I’m not very advanced at topology.



I would like to understand this perspective better. Could you give me a broad overview of how a continuous map $f:Xto Y$ “tells us something about $X$ and $Y$”?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Also, of course that if you can be embedded into a totally disconnected space, you have to be totally disconnected, and if you are a quotient of a connected space, then you are connected, etc. This is a bit too broad and ambiguous, I feel.
    $endgroup$
    – Asaf Karagila
    22 hours ago










  • $begingroup$
    Are you willing to consider the set of all continuous maps $X to Y$? If $X$ is a point, then the set of all such maps is in bijection with $Y$, and if you topologize the set of such maps appropriately (compact open topology), it is homeomorphic to $Y$. So if you could understand $text{Map}(X,Y)$ for all $X$ and $Y$, you would understand quite a bit.
    $endgroup$
    – John Palmieri
    18 hours ago












  • $begingroup$
    @JohnPalmieri yes I see that, but I am thinking about: what does a single continuous map between $X$ and $Y$ tell us, or at least an managable amount of continuous maps. A single group homomorphism tells you a lot about the two groups, and you don’t need an uncountable amount of group homomorphisms before you start to learn something.
    $endgroup$
    – user56834
    6 hours ago










  • $begingroup$
    @user56834 Between any two groups $G$ and $H$ you have a trivial morphism $gmapsto 1_H$. What does this single group homomorphism tell you about $G$ and $H$?
    $endgroup$
    – Arnaud D.
    3 mins ago










  • $begingroup$
    @ArnaudD. It tells us nothing. That’s why it’s called trivial.
    $endgroup$
    – user56834
    1 min ago
















1












$begingroup$


Category theory, roughly, is about the relation between structures of a certain kind, rather than their internal structure.



In the category of groups, I can see how this is the case: the existence of a particular group homomorphism between $G$ and $H$ really tells you alot about $G$ and $H$ as structures.



But I’ve never thought of continuous maps between topological spaces $X$ and $Y$ as “telling us something about $X$ and $Y$”. Admittedly, I’m not very advanced at topology.



I would like to understand this perspective better. Could you give me a broad overview of how a continuous map $f:Xto Y$ “tells us something about $X$ and $Y$”?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Also, of course that if you can be embedded into a totally disconnected space, you have to be totally disconnected, and if you are a quotient of a connected space, then you are connected, etc. This is a bit too broad and ambiguous, I feel.
    $endgroup$
    – Asaf Karagila
    22 hours ago










  • $begingroup$
    Are you willing to consider the set of all continuous maps $X to Y$? If $X$ is a point, then the set of all such maps is in bijection with $Y$, and if you topologize the set of such maps appropriately (compact open topology), it is homeomorphic to $Y$. So if you could understand $text{Map}(X,Y)$ for all $X$ and $Y$, you would understand quite a bit.
    $endgroup$
    – John Palmieri
    18 hours ago












  • $begingroup$
    @JohnPalmieri yes I see that, but I am thinking about: what does a single continuous map between $X$ and $Y$ tell us, or at least an managable amount of continuous maps. A single group homomorphism tells you a lot about the two groups, and you don’t need an uncountable amount of group homomorphisms before you start to learn something.
    $endgroup$
    – user56834
    6 hours ago










  • $begingroup$
    @user56834 Between any two groups $G$ and $H$ you have a trivial morphism $gmapsto 1_H$. What does this single group homomorphism tell you about $G$ and $H$?
    $endgroup$
    – Arnaud D.
    3 mins ago










  • $begingroup$
    @ArnaudD. It tells us nothing. That’s why it’s called trivial.
    $endgroup$
    – user56834
    1 min ago














1












1








1





$begingroup$


Category theory, roughly, is about the relation between structures of a certain kind, rather than their internal structure.



In the category of groups, I can see how this is the case: the existence of a particular group homomorphism between $G$ and $H$ really tells you alot about $G$ and $H$ as structures.



But I’ve never thought of continuous maps between topological spaces $X$ and $Y$ as “telling us something about $X$ and $Y$”. Admittedly, I’m not very advanced at topology.



I would like to understand this perspective better. Could you give me a broad overview of how a continuous map $f:Xto Y$ “tells us something about $X$ and $Y$”?










share|cite|improve this question









$endgroup$




Category theory, roughly, is about the relation between structures of a certain kind, rather than their internal structure.



In the category of groups, I can see how this is the case: the existence of a particular group homomorphism between $G$ and $H$ really tells you alot about $G$ and $H$ as structures.



But I’ve never thought of continuous maps between topological spaces $X$ and $Y$ as “telling us something about $X$ and $Y$”. Admittedly, I’m not very advanced at topology.



I would like to understand this perspective better. Could you give me a broad overview of how a continuous map $f:Xto Y$ “tells us something about $X$ and $Y$”?







general-topology group-theory category-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 23 hours ago









user56834user56834

3,36821252




3,36821252












  • $begingroup$
    Also, of course that if you can be embedded into a totally disconnected space, you have to be totally disconnected, and if you are a quotient of a connected space, then you are connected, etc. This is a bit too broad and ambiguous, I feel.
    $endgroup$
    – Asaf Karagila
    22 hours ago










  • $begingroup$
    Are you willing to consider the set of all continuous maps $X to Y$? If $X$ is a point, then the set of all such maps is in bijection with $Y$, and if you topologize the set of such maps appropriately (compact open topology), it is homeomorphic to $Y$. So if you could understand $text{Map}(X,Y)$ for all $X$ and $Y$, you would understand quite a bit.
    $endgroup$
    – John Palmieri
    18 hours ago












  • $begingroup$
    @JohnPalmieri yes I see that, but I am thinking about: what does a single continuous map between $X$ and $Y$ tell us, or at least an managable amount of continuous maps. A single group homomorphism tells you a lot about the two groups, and you don’t need an uncountable amount of group homomorphisms before you start to learn something.
    $endgroup$
    – user56834
    6 hours ago










  • $begingroup$
    @user56834 Between any two groups $G$ and $H$ you have a trivial morphism $gmapsto 1_H$. What does this single group homomorphism tell you about $G$ and $H$?
    $endgroup$
    – Arnaud D.
    3 mins ago










  • $begingroup$
    @ArnaudD. It tells us nothing. That’s why it’s called trivial.
    $endgroup$
    – user56834
    1 min ago


















  • $begingroup$
    Also, of course that if you can be embedded into a totally disconnected space, you have to be totally disconnected, and if you are a quotient of a connected space, then you are connected, etc. This is a bit too broad and ambiguous, I feel.
    $endgroup$
    – Asaf Karagila
    22 hours ago










  • $begingroup$
    Are you willing to consider the set of all continuous maps $X to Y$? If $X$ is a point, then the set of all such maps is in bijection with $Y$, and if you topologize the set of such maps appropriately (compact open topology), it is homeomorphic to $Y$. So if you could understand $text{Map}(X,Y)$ for all $X$ and $Y$, you would understand quite a bit.
    $endgroup$
    – John Palmieri
    18 hours ago












  • $begingroup$
    @JohnPalmieri yes I see that, but I am thinking about: what does a single continuous map between $X$ and $Y$ tell us, or at least an managable amount of continuous maps. A single group homomorphism tells you a lot about the two groups, and you don’t need an uncountable amount of group homomorphisms before you start to learn something.
    $endgroup$
    – user56834
    6 hours ago










  • $begingroup$
    @user56834 Between any two groups $G$ and $H$ you have a trivial morphism $gmapsto 1_H$. What does this single group homomorphism tell you about $G$ and $H$?
    $endgroup$
    – Arnaud D.
    3 mins ago










  • $begingroup$
    @ArnaudD. It tells us nothing. That’s why it’s called trivial.
    $endgroup$
    – user56834
    1 min ago
















$begingroup$
Also, of course that if you can be embedded into a totally disconnected space, you have to be totally disconnected, and if you are a quotient of a connected space, then you are connected, etc. This is a bit too broad and ambiguous, I feel.
$endgroup$
– Asaf Karagila
22 hours ago




$begingroup$
Also, of course that if you can be embedded into a totally disconnected space, you have to be totally disconnected, and if you are a quotient of a connected space, then you are connected, etc. This is a bit too broad and ambiguous, I feel.
$endgroup$
– Asaf Karagila
22 hours ago












$begingroup$
Are you willing to consider the set of all continuous maps $X to Y$? If $X$ is a point, then the set of all such maps is in bijection with $Y$, and if you topologize the set of such maps appropriately (compact open topology), it is homeomorphic to $Y$. So if you could understand $text{Map}(X,Y)$ for all $X$ and $Y$, you would understand quite a bit.
$endgroup$
– John Palmieri
18 hours ago






$begingroup$
Are you willing to consider the set of all continuous maps $X to Y$? If $X$ is a point, then the set of all such maps is in bijection with $Y$, and if you topologize the set of such maps appropriately (compact open topology), it is homeomorphic to $Y$. So if you could understand $text{Map}(X,Y)$ for all $X$ and $Y$, you would understand quite a bit.
$endgroup$
– John Palmieri
18 hours ago














$begingroup$
@JohnPalmieri yes I see that, but I am thinking about: what does a single continuous map between $X$ and $Y$ tell us, or at least an managable amount of continuous maps. A single group homomorphism tells you a lot about the two groups, and you don’t need an uncountable amount of group homomorphisms before you start to learn something.
$endgroup$
– user56834
6 hours ago




$begingroup$
@JohnPalmieri yes I see that, but I am thinking about: what does a single continuous map between $X$ and $Y$ tell us, or at least an managable amount of continuous maps. A single group homomorphism tells you a lot about the two groups, and you don’t need an uncountable amount of group homomorphisms before you start to learn something.
$endgroup$
– user56834
6 hours ago












$begingroup$
@user56834 Between any two groups $G$ and $H$ you have a trivial morphism $gmapsto 1_H$. What does this single group homomorphism tell you about $G$ and $H$?
$endgroup$
– Arnaud D.
3 mins ago




$begingroup$
@user56834 Between any two groups $G$ and $H$ you have a trivial morphism $gmapsto 1_H$. What does this single group homomorphism tell you about $G$ and $H$?
$endgroup$
– Arnaud D.
3 mins ago












$begingroup$
@ArnaudD. It tells us nothing. That’s why it’s called trivial.
$endgroup$
– user56834
1 min ago




$begingroup$
@ArnaudD. It tells us nothing. That’s why it’s called trivial.
$endgroup$
– user56834
1 min ago










2 Answers
2






active

oldest

votes


















1












$begingroup$

In my opinion the easiest way to understand it is to look at the isomorphisms, instead of the continous maps, because they tell much more about the required structure on $X$ and $Y$. If you continue on your example with groups, two isomorphic groups $X$ and $Y$ are such that they will have exactly the same structure for everything regarding groups.



Now it turns out that for topological spaces, the thing that we need to say "$X$ and $Y$ are exactly the same for everything regarding the topology" is the notion of homeomorphism, that is a bijective continuous map whose inverse is continuous. It is not at first obvious why this is the right notion, so let me explain. A topology on $X$ is a class $tau_Xsubsetmathcal{P}X$ of open sets in $X$, satisfying some properties. Now take two topological spaces $X$ and $Y$ with a homeomorphism $f : X to Y$. $f$ is bijective means that $X$ and $Y$ are the same as sets, $f$ is continuous means that the preimage of $mathcal{O}_Yintau_Y$ is in $tau_X$, and $f^{-1}$ is injective means that the preimage of $mathcal{O}_Xintau_X$ is in $tau_Y$. If you put all this together, it says that $f$ is a bijection between $X$ and $Y$, and the bijection induced by $f$ between $mathcal{P}X$ and $mathcal{P}Y$ can restrict to bijection between $tau_X$ and $tau_Y$. And this formulation relates much more to the definition of topological space.



Now if $f$ is a continuous map between $X$ and $Y$, by definition, the preimage of every open of $Y$ is open in $X$, so this is already "telling us something relating the topological structure on $X$ with the one on $Y$", but I admit it is pretty vague.



To my knowledge, if you want more general properties to hold, you will want some extra properties for your topological (like separation...), because the category of topological spaces is very weird, but I might be wrong on that point.



Now let's assume that you have a category of "nice" topological spaces (whatever nice means). Studying topology is hard, but we can compute its fundamental group. Now a continuous map $f : X to Y$ induces a map $pi_1(f) : pi_1(X)topi_1(Y)$, which happens to be a group homomorphism, so if you already know how a group homomorphism tells us something about the similarity between the two groups, and if you know (or accept), that the fundamental group tells us a lot about the topology of the spaces, then you see indirectly how a continuous maps tells indeed a lot about the topology of the two spaces.



Note that this explanation I gave is very weak, in the sense that a continuous map tells us a lot more than a group homomorphism on the fundamental groups, it induces also a morphism between the higher homotopy groups (and also teh groupoid version of this statements, for non-connected spaces). So in the end it really says quite a lot, but not much that I can give explicitly






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    I think it is kind of a philosophy of understanding what is a space. Let's start with the simplest case.



    We know that what is a vector space $V$ over some field $k$. And in linear algebra, we concentrate on the linear properties of a linear space. However, the linear properties are those preversed by linear maps, so the set of all linear functions $mathrm{Hom}(V,k)$ is crucial. But it isn't hard to show that $Vcongmathrm{Hom}(V,k)$, which means if we know all the linear functions, we already know the linear structure of our original vector space $V$.



    Move to a senario a little bit complicated. Suppose $mathbb{A}_k^n$ be the affine space associated with the vector space $k^n$. Over this space $mathbb{A}_k^n$, we want to know the affine properties preserved by affine maps. Again, if we know all the affine functions (from $mathbb{A}_k^n$ to $k$), we can construct the original space $mathbb{A}_k^n$.



    If we keep moving on, to somewhere called differential geometry, things become more clear. We have some coordinate patches, then all functions on the manifold can be (and should be) dealt with by taking the pull-back (locally). Just an example, the cotangent space of a manifold at some point is defined by something on the set of all smooth functions, and its dimension coincide with the dimension of the manifold. You see this is a geometrical property that we can extract if we only know all the functions from a manifold to $mathbb{R}$.



    All the cases are saying, that as long as we know all the functions form some object in our category, to some really simple object in our category, we can always know many properties preserved by the morphisms in this category, even constructing the object again.



    Grothendick said something like in the classical algebraic geometries, the way we studied varieties was wrong. This was because varieties should not be defined by the zero locus of some polynomials. Almost all the properties of a space come from studying the set of all possible good functions, so the definition of a space should also be there if we know all the functions. This is basically the scheme. We first know all functions, which form a ring $R$, then we know the space should be $|mathrm{Spec}~R|$, and the topology on this space is Zariski topology. And $R$ is really the set of functions, while locally we can have more functions.



    This pattern also shows in algebraic topology. Suppose we have a compact surface $S$, and we know all the functions from $S^1$ and $S^2$ to $S$, then we already know the fundamental group of $S$ and hence $H_1$. Therefore, we essentially know what $S$ is.



    All these do not purely come from algebraic topology, but they are all saying the same thing. In a category, if we know all the morphisms from an object to other object, then this object should be determined uniquely up to isomorphism. I hope this could be helpful.






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      2 Answers
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      2 Answers
      2






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      active

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      1












      $begingroup$

      In my opinion the easiest way to understand it is to look at the isomorphisms, instead of the continous maps, because they tell much more about the required structure on $X$ and $Y$. If you continue on your example with groups, two isomorphic groups $X$ and $Y$ are such that they will have exactly the same structure for everything regarding groups.



      Now it turns out that for topological spaces, the thing that we need to say "$X$ and $Y$ are exactly the same for everything regarding the topology" is the notion of homeomorphism, that is a bijective continuous map whose inverse is continuous. It is not at first obvious why this is the right notion, so let me explain. A topology on $X$ is a class $tau_Xsubsetmathcal{P}X$ of open sets in $X$, satisfying some properties. Now take two topological spaces $X$ and $Y$ with a homeomorphism $f : X to Y$. $f$ is bijective means that $X$ and $Y$ are the same as sets, $f$ is continuous means that the preimage of $mathcal{O}_Yintau_Y$ is in $tau_X$, and $f^{-1}$ is injective means that the preimage of $mathcal{O}_Xintau_X$ is in $tau_Y$. If you put all this together, it says that $f$ is a bijection between $X$ and $Y$, and the bijection induced by $f$ between $mathcal{P}X$ and $mathcal{P}Y$ can restrict to bijection between $tau_X$ and $tau_Y$. And this formulation relates much more to the definition of topological space.



      Now if $f$ is a continuous map between $X$ and $Y$, by definition, the preimage of every open of $Y$ is open in $X$, so this is already "telling us something relating the topological structure on $X$ with the one on $Y$", but I admit it is pretty vague.



      To my knowledge, if you want more general properties to hold, you will want some extra properties for your topological (like separation...), because the category of topological spaces is very weird, but I might be wrong on that point.



      Now let's assume that you have a category of "nice" topological spaces (whatever nice means). Studying topology is hard, but we can compute its fundamental group. Now a continuous map $f : X to Y$ induces a map $pi_1(f) : pi_1(X)topi_1(Y)$, which happens to be a group homomorphism, so if you already know how a group homomorphism tells us something about the similarity between the two groups, and if you know (or accept), that the fundamental group tells us a lot about the topology of the spaces, then you see indirectly how a continuous maps tells indeed a lot about the topology of the two spaces.



      Note that this explanation I gave is very weak, in the sense that a continuous map tells us a lot more than a group homomorphism on the fundamental groups, it induces also a morphism between the higher homotopy groups (and also teh groupoid version of this statements, for non-connected spaces). So in the end it really says quite a lot, but not much that I can give explicitly






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        In my opinion the easiest way to understand it is to look at the isomorphisms, instead of the continous maps, because they tell much more about the required structure on $X$ and $Y$. If you continue on your example with groups, two isomorphic groups $X$ and $Y$ are such that they will have exactly the same structure for everything regarding groups.



        Now it turns out that for topological spaces, the thing that we need to say "$X$ and $Y$ are exactly the same for everything regarding the topology" is the notion of homeomorphism, that is a bijective continuous map whose inverse is continuous. It is not at first obvious why this is the right notion, so let me explain. A topology on $X$ is a class $tau_Xsubsetmathcal{P}X$ of open sets in $X$, satisfying some properties. Now take two topological spaces $X$ and $Y$ with a homeomorphism $f : X to Y$. $f$ is bijective means that $X$ and $Y$ are the same as sets, $f$ is continuous means that the preimage of $mathcal{O}_Yintau_Y$ is in $tau_X$, and $f^{-1}$ is injective means that the preimage of $mathcal{O}_Xintau_X$ is in $tau_Y$. If you put all this together, it says that $f$ is a bijection between $X$ and $Y$, and the bijection induced by $f$ between $mathcal{P}X$ and $mathcal{P}Y$ can restrict to bijection between $tau_X$ and $tau_Y$. And this formulation relates much more to the definition of topological space.



        Now if $f$ is a continuous map between $X$ and $Y$, by definition, the preimage of every open of $Y$ is open in $X$, so this is already "telling us something relating the topological structure on $X$ with the one on $Y$", but I admit it is pretty vague.



        To my knowledge, if you want more general properties to hold, you will want some extra properties for your topological (like separation...), because the category of topological spaces is very weird, but I might be wrong on that point.



        Now let's assume that you have a category of "nice" topological spaces (whatever nice means). Studying topology is hard, but we can compute its fundamental group. Now a continuous map $f : X to Y$ induces a map $pi_1(f) : pi_1(X)topi_1(Y)$, which happens to be a group homomorphism, so if you already know how a group homomorphism tells us something about the similarity between the two groups, and if you know (or accept), that the fundamental group tells us a lot about the topology of the spaces, then you see indirectly how a continuous maps tells indeed a lot about the topology of the two spaces.



        Note that this explanation I gave is very weak, in the sense that a continuous map tells us a lot more than a group homomorphism on the fundamental groups, it induces also a morphism between the higher homotopy groups (and also teh groupoid version of this statements, for non-connected spaces). So in the end it really says quite a lot, but not much that I can give explicitly






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          In my opinion the easiest way to understand it is to look at the isomorphisms, instead of the continous maps, because they tell much more about the required structure on $X$ and $Y$. If you continue on your example with groups, two isomorphic groups $X$ and $Y$ are such that they will have exactly the same structure for everything regarding groups.



          Now it turns out that for topological spaces, the thing that we need to say "$X$ and $Y$ are exactly the same for everything regarding the topology" is the notion of homeomorphism, that is a bijective continuous map whose inverse is continuous. It is not at first obvious why this is the right notion, so let me explain. A topology on $X$ is a class $tau_Xsubsetmathcal{P}X$ of open sets in $X$, satisfying some properties. Now take two topological spaces $X$ and $Y$ with a homeomorphism $f : X to Y$. $f$ is bijective means that $X$ and $Y$ are the same as sets, $f$ is continuous means that the preimage of $mathcal{O}_Yintau_Y$ is in $tau_X$, and $f^{-1}$ is injective means that the preimage of $mathcal{O}_Xintau_X$ is in $tau_Y$. If you put all this together, it says that $f$ is a bijection between $X$ and $Y$, and the bijection induced by $f$ between $mathcal{P}X$ and $mathcal{P}Y$ can restrict to bijection between $tau_X$ and $tau_Y$. And this formulation relates much more to the definition of topological space.



          Now if $f$ is a continuous map between $X$ and $Y$, by definition, the preimage of every open of $Y$ is open in $X$, so this is already "telling us something relating the topological structure on $X$ with the one on $Y$", but I admit it is pretty vague.



          To my knowledge, if you want more general properties to hold, you will want some extra properties for your topological (like separation...), because the category of topological spaces is very weird, but I might be wrong on that point.



          Now let's assume that you have a category of "nice" topological spaces (whatever nice means). Studying topology is hard, but we can compute its fundamental group. Now a continuous map $f : X to Y$ induces a map $pi_1(f) : pi_1(X)topi_1(Y)$, which happens to be a group homomorphism, so if you already know how a group homomorphism tells us something about the similarity between the two groups, and if you know (or accept), that the fundamental group tells us a lot about the topology of the spaces, then you see indirectly how a continuous maps tells indeed a lot about the topology of the two spaces.



          Note that this explanation I gave is very weak, in the sense that a continuous map tells us a lot more than a group homomorphism on the fundamental groups, it induces also a morphism between the higher homotopy groups (and also teh groupoid version of this statements, for non-connected spaces). So in the end it really says quite a lot, but not much that I can give explicitly






          share|cite|improve this answer









          $endgroup$



          In my opinion the easiest way to understand it is to look at the isomorphisms, instead of the continous maps, because they tell much more about the required structure on $X$ and $Y$. If you continue on your example with groups, two isomorphic groups $X$ and $Y$ are such that they will have exactly the same structure for everything regarding groups.



          Now it turns out that for topological spaces, the thing that we need to say "$X$ and $Y$ are exactly the same for everything regarding the topology" is the notion of homeomorphism, that is a bijective continuous map whose inverse is continuous. It is not at first obvious why this is the right notion, so let me explain. A topology on $X$ is a class $tau_Xsubsetmathcal{P}X$ of open sets in $X$, satisfying some properties. Now take two topological spaces $X$ and $Y$ with a homeomorphism $f : X to Y$. $f$ is bijective means that $X$ and $Y$ are the same as sets, $f$ is continuous means that the preimage of $mathcal{O}_Yintau_Y$ is in $tau_X$, and $f^{-1}$ is injective means that the preimage of $mathcal{O}_Xintau_X$ is in $tau_Y$. If you put all this together, it says that $f$ is a bijection between $X$ and $Y$, and the bijection induced by $f$ between $mathcal{P}X$ and $mathcal{P}Y$ can restrict to bijection between $tau_X$ and $tau_Y$. And this formulation relates much more to the definition of topological space.



          Now if $f$ is a continuous map between $X$ and $Y$, by definition, the preimage of every open of $Y$ is open in $X$, so this is already "telling us something relating the topological structure on $X$ with the one on $Y$", but I admit it is pretty vague.



          To my knowledge, if you want more general properties to hold, you will want some extra properties for your topological (like separation...), because the category of topological spaces is very weird, but I might be wrong on that point.



          Now let's assume that you have a category of "nice" topological spaces (whatever nice means). Studying topology is hard, but we can compute its fundamental group. Now a continuous map $f : X to Y$ induces a map $pi_1(f) : pi_1(X)topi_1(Y)$, which happens to be a group homomorphism, so if you already know how a group homomorphism tells us something about the similarity between the two groups, and if you know (or accept), that the fundamental group tells us a lot about the topology of the spaces, then you see indirectly how a continuous maps tells indeed a lot about the topology of the two spaces.



          Note that this explanation I gave is very weak, in the sense that a continuous map tells us a lot more than a group homomorphism on the fundamental groups, it induces also a morphism between the higher homotopy groups (and also teh groupoid version of this statements, for non-connected spaces). So in the end it really says quite a lot, but not much that I can give explicitly







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 21 hours ago









          Thibaut BenjaminThibaut Benjamin

          1667




          1667























              0












              $begingroup$

              I think it is kind of a philosophy of understanding what is a space. Let's start with the simplest case.



              We know that what is a vector space $V$ over some field $k$. And in linear algebra, we concentrate on the linear properties of a linear space. However, the linear properties are those preversed by linear maps, so the set of all linear functions $mathrm{Hom}(V,k)$ is crucial. But it isn't hard to show that $Vcongmathrm{Hom}(V,k)$, which means if we know all the linear functions, we already know the linear structure of our original vector space $V$.



              Move to a senario a little bit complicated. Suppose $mathbb{A}_k^n$ be the affine space associated with the vector space $k^n$. Over this space $mathbb{A}_k^n$, we want to know the affine properties preserved by affine maps. Again, if we know all the affine functions (from $mathbb{A}_k^n$ to $k$), we can construct the original space $mathbb{A}_k^n$.



              If we keep moving on, to somewhere called differential geometry, things become more clear. We have some coordinate patches, then all functions on the manifold can be (and should be) dealt with by taking the pull-back (locally). Just an example, the cotangent space of a manifold at some point is defined by something on the set of all smooth functions, and its dimension coincide with the dimension of the manifold. You see this is a geometrical property that we can extract if we only know all the functions from a manifold to $mathbb{R}$.



              All the cases are saying, that as long as we know all the functions form some object in our category, to some really simple object in our category, we can always know many properties preserved by the morphisms in this category, even constructing the object again.



              Grothendick said something like in the classical algebraic geometries, the way we studied varieties was wrong. This was because varieties should not be defined by the zero locus of some polynomials. Almost all the properties of a space come from studying the set of all possible good functions, so the definition of a space should also be there if we know all the functions. This is basically the scheme. We first know all functions, which form a ring $R$, then we know the space should be $|mathrm{Spec}~R|$, and the topology on this space is Zariski topology. And $R$ is really the set of functions, while locally we can have more functions.



              This pattern also shows in algebraic topology. Suppose we have a compact surface $S$, and we know all the functions from $S^1$ and $S^2$ to $S$, then we already know the fundamental group of $S$ and hence $H_1$. Therefore, we essentially know what $S$ is.



              All these do not purely come from algebraic topology, but they are all saying the same thing. In a category, if we know all the morphisms from an object to other object, then this object should be determined uniquely up to isomorphism. I hope this could be helpful.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                I think it is kind of a philosophy of understanding what is a space. Let's start with the simplest case.



                We know that what is a vector space $V$ over some field $k$. And in linear algebra, we concentrate on the linear properties of a linear space. However, the linear properties are those preversed by linear maps, so the set of all linear functions $mathrm{Hom}(V,k)$ is crucial. But it isn't hard to show that $Vcongmathrm{Hom}(V,k)$, which means if we know all the linear functions, we already know the linear structure of our original vector space $V$.



                Move to a senario a little bit complicated. Suppose $mathbb{A}_k^n$ be the affine space associated with the vector space $k^n$. Over this space $mathbb{A}_k^n$, we want to know the affine properties preserved by affine maps. Again, if we know all the affine functions (from $mathbb{A}_k^n$ to $k$), we can construct the original space $mathbb{A}_k^n$.



                If we keep moving on, to somewhere called differential geometry, things become more clear. We have some coordinate patches, then all functions on the manifold can be (and should be) dealt with by taking the pull-back (locally). Just an example, the cotangent space of a manifold at some point is defined by something on the set of all smooth functions, and its dimension coincide with the dimension of the manifold. You see this is a geometrical property that we can extract if we only know all the functions from a manifold to $mathbb{R}$.



                All the cases are saying, that as long as we know all the functions form some object in our category, to some really simple object in our category, we can always know many properties preserved by the morphisms in this category, even constructing the object again.



                Grothendick said something like in the classical algebraic geometries, the way we studied varieties was wrong. This was because varieties should not be defined by the zero locus of some polynomials. Almost all the properties of a space come from studying the set of all possible good functions, so the definition of a space should also be there if we know all the functions. This is basically the scheme. We first know all functions, which form a ring $R$, then we know the space should be $|mathrm{Spec}~R|$, and the topology on this space is Zariski topology. And $R$ is really the set of functions, while locally we can have more functions.



                This pattern also shows in algebraic topology. Suppose we have a compact surface $S$, and we know all the functions from $S^1$ and $S^2$ to $S$, then we already know the fundamental group of $S$ and hence $H_1$. Therefore, we essentially know what $S$ is.



                All these do not purely come from algebraic topology, but they are all saying the same thing. In a category, if we know all the morphisms from an object to other object, then this object should be determined uniquely up to isomorphism. I hope this could be helpful.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I think it is kind of a philosophy of understanding what is a space. Let's start with the simplest case.



                  We know that what is a vector space $V$ over some field $k$. And in linear algebra, we concentrate on the linear properties of a linear space. However, the linear properties are those preversed by linear maps, so the set of all linear functions $mathrm{Hom}(V,k)$ is crucial. But it isn't hard to show that $Vcongmathrm{Hom}(V,k)$, which means if we know all the linear functions, we already know the linear structure of our original vector space $V$.



                  Move to a senario a little bit complicated. Suppose $mathbb{A}_k^n$ be the affine space associated with the vector space $k^n$. Over this space $mathbb{A}_k^n$, we want to know the affine properties preserved by affine maps. Again, if we know all the affine functions (from $mathbb{A}_k^n$ to $k$), we can construct the original space $mathbb{A}_k^n$.



                  If we keep moving on, to somewhere called differential geometry, things become more clear. We have some coordinate patches, then all functions on the manifold can be (and should be) dealt with by taking the pull-back (locally). Just an example, the cotangent space of a manifold at some point is defined by something on the set of all smooth functions, and its dimension coincide with the dimension of the manifold. You see this is a geometrical property that we can extract if we only know all the functions from a manifold to $mathbb{R}$.



                  All the cases are saying, that as long as we know all the functions form some object in our category, to some really simple object in our category, we can always know many properties preserved by the morphisms in this category, even constructing the object again.



                  Grothendick said something like in the classical algebraic geometries, the way we studied varieties was wrong. This was because varieties should not be defined by the zero locus of some polynomials. Almost all the properties of a space come from studying the set of all possible good functions, so the definition of a space should also be there if we know all the functions. This is basically the scheme. We first know all functions, which form a ring $R$, then we know the space should be $|mathrm{Spec}~R|$, and the topology on this space is Zariski topology. And $R$ is really the set of functions, while locally we can have more functions.



                  This pattern also shows in algebraic topology. Suppose we have a compact surface $S$, and we know all the functions from $S^1$ and $S^2$ to $S$, then we already know the fundamental group of $S$ and hence $H_1$. Therefore, we essentially know what $S$ is.



                  All these do not purely come from algebraic topology, but they are all saying the same thing. In a category, if we know all the morphisms from an object to other object, then this object should be determined uniquely up to isomorphism. I hope this could be helpful.






                  share|cite|improve this answer









                  $endgroup$



                  I think it is kind of a philosophy of understanding what is a space. Let's start with the simplest case.



                  We know that what is a vector space $V$ over some field $k$. And in linear algebra, we concentrate on the linear properties of a linear space. However, the linear properties are those preversed by linear maps, so the set of all linear functions $mathrm{Hom}(V,k)$ is crucial. But it isn't hard to show that $Vcongmathrm{Hom}(V,k)$, which means if we know all the linear functions, we already know the linear structure of our original vector space $V$.



                  Move to a senario a little bit complicated. Suppose $mathbb{A}_k^n$ be the affine space associated with the vector space $k^n$. Over this space $mathbb{A}_k^n$, we want to know the affine properties preserved by affine maps. Again, if we know all the affine functions (from $mathbb{A}_k^n$ to $k$), we can construct the original space $mathbb{A}_k^n$.



                  If we keep moving on, to somewhere called differential geometry, things become more clear. We have some coordinate patches, then all functions on the manifold can be (and should be) dealt with by taking the pull-back (locally). Just an example, the cotangent space of a manifold at some point is defined by something on the set of all smooth functions, and its dimension coincide with the dimension of the manifold. You see this is a geometrical property that we can extract if we only know all the functions from a manifold to $mathbb{R}$.



                  All the cases are saying, that as long as we know all the functions form some object in our category, to some really simple object in our category, we can always know many properties preserved by the morphisms in this category, even constructing the object again.



                  Grothendick said something like in the classical algebraic geometries, the way we studied varieties was wrong. This was because varieties should not be defined by the zero locus of some polynomials. Almost all the properties of a space come from studying the set of all possible good functions, so the definition of a space should also be there if we know all the functions. This is basically the scheme. We first know all functions, which form a ring $R$, then we know the space should be $|mathrm{Spec}~R|$, and the topology on this space is Zariski topology. And $R$ is really the set of functions, while locally we can have more functions.



                  This pattern also shows in algebraic topology. Suppose we have a compact surface $S$, and we know all the functions from $S^1$ and $S^2$ to $S$, then we already know the fundamental group of $S$ and hence $H_1$. Therefore, we essentially know what $S$ is.



                  All these do not purely come from algebraic topology, but they are all saying the same thing. In a category, if we know all the morphisms from an object to other object, then this object should be determined uniquely up to isomorphism. I hope this could be helpful.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 20 hours ago









                  Guanyu LiGuanyu Li

                  144




                  144






























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