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0

$begingroup$

I'm familiar with the universal property of coproducts, but I have no idea how to prove something like $h circ [ f, g ] equiv [ h circ f, h circ g ]$. The diagram obviously commutes, but I can't see how you'd prove this with the category + coproduct identities.

How might I go about this?

category-theory

share|cite|improve this question

edited 2 days ago

lightandlight

asked 2 days ago

lightandlightlightandlight

1113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What does $langle f + g rangle$ mean?
  $endgroup$
  – Qiaochu Yuan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's the copairing of f and g / eliminator for the coproduct. The unique arrow given by the universal property.
  $endgroup$
  – lightandlight
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You mean, $f, g$ are two morphisms $f : a to c, g : b to c$, and $langle f + g rangle$ is the morphism $a sqcup b to c$ whose components are $f$ and $g$?
  $endgroup$
  – Qiaochu Yuan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yep, that's right
  $endgroup$
  – lightandlight
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The notation I've usually seen for that is $[f,g]$. $f+g$ is the bifunctorial action, and $langle f,grangle$ is commonly used for the pairing map for products. I've never seen $langle f+grangle$.
  $endgroup$
  – Derek Elkins
  2 days ago
  
  
  

  
  
  
  

 | 
show 3 more comments

0

$begingroup$

I'm familiar with the universal property of coproducts, but I have no idea how to prove something like $h circ [ f, g ] equiv [ h circ f, h circ g ]$. The diagram obviously commutes, but I can't see how you'd prove this with the category + coproduct identities.

How might I go about this?

category-theory

share|cite|improve this question

edited 2 days ago

lightandlight

asked 2 days ago

lightandlightlightandlight

1113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What does $langle f + g rangle$ mean?
  $endgroup$
  – Qiaochu Yuan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's the copairing of f and g / eliminator for the coproduct. The unique arrow given by the universal property.
  $endgroup$
  – lightandlight
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You mean, $f, g$ are two morphisms $f : a to c, g : b to c$, and $langle f + g rangle$ is the morphism $a sqcup b to c$ whose components are $f$ and $g$?
  $endgroup$
  – Qiaochu Yuan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yep, that's right
  $endgroup$
  – lightandlight
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The notation I've usually seen for that is $[f,g]$. $f+g$ is the bifunctorial action, and $langle f,grangle$ is commonly used for the pairing map for products. I've never seen $langle f+grangle$.
  $endgroup$
  – Derek Elkins
  2 days ago
  
  
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

I'm familiar with the universal property of coproducts, but I have no idea how to prove something like $h circ [ f, g ] equiv [ h circ f, h circ g ]$. The diagram obviously commutes, but I can't see how you'd prove this with the category + coproduct identities.

How might I go about this?

category-theory

share|cite|improve this question

edited 2 days ago

lightandlight

asked 2 days ago

lightandlightlightandlight

1113

$endgroup$

share|cite|improve this question

edited 2 days ago

lightandlight

asked 2 days ago

lightandlightlightandlight

1113

share|cite|improve this question

edited 2 days ago

lightandlight

asked 2 days ago

lightandlightlightandlight

1113

asked 2 days ago

lightandlightlightandlight

1113

asked 2 days ago

lightandlightlightandlight

1113