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$begingroup$

First, notice that in the diagram you provided, if you are given two points $p ne q$ in the square, those two points $p,q$ are identified under the gluing if and only if $p,q$ are endpoints of a "diameter" of the square, meaning that the line segment $overline{pq}$ passes through the center of the square.

Now round out the square to get a round disc $D^2 = {(x,y) mid x^2 + y^2 le 1}$. It's boundary circle is $partial D^2 = {(x,y) mid x^2 + y^2 = 1$. Again the identifications are that two points $p ne q in D^2$ are identified if and only if $overline{pq}$ is a diameter of $D^2$.

Now consider the upper hemisphere of $S^2$, namely
$$S^2_+ = {(x,y,sqrt{1-x^2-y^2}) mid x^2 + y^2 le 1}
$$
The boundary of $S^2_+$ is the equator of $S^2$. Again two points $p ne q in S^2_+$ are identified if and only if $p,q$ endpoints of a diameter of the equator.

Finally, take the union of the upper and lower hemispheres to get the whole sphere $S^2$, where two points $p ne q in S^2$ are identified if and only if they are endpoints of a diameter of $S^2$. You get a bijection between the quotient of $S^2_+$ and the quotient of $S^2$, because if $p ne q in S^2_+$ then they are identified in the quotient of $S^2_+$ if and only if they are identified in the quotient of $S^2$.

That should give you enough intuition. There are some further formalities that require some topology, to prove that this sequence of maps between quotients is a sequence of homeomorphisms.

share|cite|improve this answer

answered Feb 15 at 20:22

Lee MosherLee Mosher

51.8k33889

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm referring to some universality theorems satisfied by quotient maps, which one can find for example in Munkres "Topology".
  $endgroup$
  – Lee Mosher
  Feb 16 at 14:42
  

  
  
  
  

add a comment | 

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1

$begingroup$

First, notice that in the diagram you provided, if you are given two points $p ne q$ in the square, those two points $p,q$ are identified under the gluing if and only if $p,q$ are endpoints of a "diameter" of the square, meaning that the line segment $overline{pq}$ passes through the center of the square.

Now round out the square to get a round disc $D^2 = {(x,y) mid x^2 + y^2 le 1}$. It's boundary circle is $partial D^2 = {(x,y) mid x^2 + y^2 = 1$. Again the identifications are that two points $p ne q in D^2$ are identified if and only if $overline{pq}$ is a diameter of $D^2$.

Now consider the upper hemisphere of $S^2$, namely
$$S^2_+ = {(x,y,sqrt{1-x^2-y^2}) mid x^2 + y^2 le 1}
$$
The boundary of $S^2_+$ is the equator of $S^2$. Again two points $p ne q in S^2_+$ are identified if and only if $p,q$ endpoints of a diameter of the equator.

Finally, take the union of the upper and lower hemispheres to get the whole sphere $S^2$, where two points $p ne q in S^2$ are identified if and only if they are endpoints of a diameter of $S^2$. You get a bijection between the quotient of $S^2_+$ and the quotient of $S^2$, because if $p ne q in S^2_+$ then they are identified in the quotient of $S^2_+$ if and only if they are identified in the quotient of $S^2$.

That should give you enough intuition. There are some further formalities that require some topology, to prove that this sequence of maps between quotients is a sequence of homeomorphisms.

share|cite|improve this answer

answered Feb 15 at 20:22

Lee MosherLee Mosher

51.8k33889

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm referring to some universality theorems satisfied by quotient maps, which one can find for example in Munkres "Topology".
  $endgroup$
  – Lee Mosher
  Feb 16 at 14:42
  

  
  
  
  

add a comment | 

1

$begingroup$

First, notice that in the diagram you provided, if you are given two points $p ne q$ in the square, those two points $p,q$ are identified under the gluing if and only if $p,q$ are endpoints of a "diameter" of the square, meaning that the line segment $overline{pq}$ passes through the center of the square.

Now round out the square to get a round disc $D^2 = {(x,y) mid x^2 + y^2 le 1}$. It's boundary circle is $partial D^2 = {(x,y) mid x^2 + y^2 = 1$. Again the identifications are that two points $p ne q in D^2$ are identified if and only if $overline{pq}$ is a diameter of $D^2$.

Now consider the upper hemisphere of $S^2$, namely
$$S^2_+ = {(x,y,sqrt{1-x^2-y^2}) mid x^2 + y^2 le 1}
$$
The boundary of $S^2_+$ is the equator of $S^2$. Again two points $p ne q in S^2_+$ are identified if and only if $p,q$ endpoints of a diameter of the equator.

Finally, take the union of the upper and lower hemispheres to get the whole sphere $S^2$, where two points $p ne q in S^2$ are identified if and only if they are endpoints of a diameter of $S^2$. You get a bijection between the quotient of $S^2_+$ and the quotient of $S^2$, because if $p ne q in S^2_+$ then they are identified in the quotient of $S^2_+$ if and only if they are identified in the quotient of $S^2$.

That should give you enough intuition. There are some further formalities that require some topology, to prove that this sequence of maps between quotients is a sequence of homeomorphisms.

share|cite|improve this answer

answered Feb 15 at 20:22

Lee MosherLee Mosher

51.8k33889

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm referring to some universality theorems satisfied by quotient maps, which one can find for example in Munkres "Topology".
  $endgroup$
  – Lee Mosher
  Feb 16 at 14:42
  

  
  
  
  

add a comment | 

$begingroup$

First, notice that in the diagram you provided, if you are given two points $p ne q$ in the square, those two points $p,q$ are identified under the gluing if and only if $p,q$ are endpoints of a "diameter" of the square, meaning that the line segment $overline{pq}$ passes through the center of the square.

Now round out the square to get a round disc $D^2 = {(x,y) mid x^2 + y^2 le 1}$. It's boundary circle is $partial D^2 = {(x,y) mid x^2 + y^2 = 1$. Again the identifications are that two points $p ne q in D^2$ are identified if and only if $overline{pq}$ is a diameter of $D^2$.

Now consider the upper hemisphere of $S^2$, namely
$$S^2_+ = {(x,y,sqrt{1-x^2-y^2}) mid x^2 + y^2 le 1}
$$
The boundary of $S^2_+$ is the equator of $S^2$. Again two points $p ne q in S^2_+$ are identified if and only if $p,q$ endpoints of a diameter of the equator.

Finally, take the union of the upper and lower hemispheres to get the whole sphere $S^2$, where two points $p ne q in S^2$ are identified if and only if they are endpoints of a diameter of $S^2$. You get a bijection between the quotient of $S^2_+$ and the quotient of $S^2$, because if $p ne q in S^2_+$ then they are identified in the quotient of $S^2_+$ if and only if they are identified in the quotient of $S^2$.

That should give you enough intuition. There are some further formalities that require some topology, to prove that this sequence of maps between quotients is a sequence of homeomorphisms.

share|cite|improve this answer

answered Feb 15 at 20:22

Lee MosherLee Mosher

51.8k33889

$endgroup$

share|cite|improve this answer

answered Feb 15 at 20:22

Lee MosherLee Mosher

51.8k33889

answered Feb 15 at 20:22

Lee MosherLee Mosher

51.8k33889

answered Feb 15 at 20:22

Lee MosherLee Mosher

51.8k33889