Struggling to understand why $Hom(-, -)$ is a bifunctor and how exactly it suppose to map?












1












$begingroup$


Seems like I do understand covariant the $Hom(A, -)$ and contravariant $Hom(-, A)$ functors. But when it comes down to the $Hom(-, -)$, then I am getting lost:



1) How does it work? What does it do with objects and how does it treat morphisms? [I assume it takes a product category $C times C$ as an input, but still underlaying details are unclear: I can't realize what is fixed and what is free.]



2) Why is it common to state that $Hom(-, -)$ is a kind of $C^{op} times C mapsto mathbb{SET}$ functor? [My assumption is that it because iterating over the first argument, having the second one fixed (i.e. $Hom(A, A'), Hom(B, A'), Hom(C, A'), ...$ naturally gives a contravariant functor, while same kind of iteration on the second argument - gives a covariant functor.]



2.1) Is it a matter of preference? In other words, is it safe to say that $C times C^{op} mapsto mathbb{SET}$ is equally valid definition? [I assume it is not matter of preference, but still better to ask.]



Could someone provide crystall clear definition of what $Hom(-, -)$ is and how should I think of it as an instance of $C times C^{op} mapsto mathbb{SET}$ functors?










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$endgroup$








  • 1




    $begingroup$
    Re: (2) it's for the reason you say. Given a pair of maps $f:Ato B$ and $g:Cto D$, $(f,g)$ is a morphism $(B,C)to(A,D)$ in $C^{op}times C$; in $C$ composition $gcirc-circ f$ gives you a function $hom(B,C)tohom(A,D)$.
    $endgroup$
    – Malice Vidrine
    Jan 11 at 18:58


















1












$begingroup$


Seems like I do understand covariant the $Hom(A, -)$ and contravariant $Hom(-, A)$ functors. But when it comes down to the $Hom(-, -)$, then I am getting lost:



1) How does it work? What does it do with objects and how does it treat morphisms? [I assume it takes a product category $C times C$ as an input, but still underlaying details are unclear: I can't realize what is fixed and what is free.]



2) Why is it common to state that $Hom(-, -)$ is a kind of $C^{op} times C mapsto mathbb{SET}$ functor? [My assumption is that it because iterating over the first argument, having the second one fixed (i.e. $Hom(A, A'), Hom(B, A'), Hom(C, A'), ...$ naturally gives a contravariant functor, while same kind of iteration on the second argument - gives a covariant functor.]



2.1) Is it a matter of preference? In other words, is it safe to say that $C times C^{op} mapsto mathbb{SET}$ is equally valid definition? [I assume it is not matter of preference, but still better to ask.]



Could someone provide crystall clear definition of what $Hom(-, -)$ is and how should I think of it as an instance of $C times C^{op} mapsto mathbb{SET}$ functors?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Re: (2) it's for the reason you say. Given a pair of maps $f:Ato B$ and $g:Cto D$, $(f,g)$ is a morphism $(B,C)to(A,D)$ in $C^{op}times C$; in $C$ composition $gcirc-circ f$ gives you a function $hom(B,C)tohom(A,D)$.
    $endgroup$
    – Malice Vidrine
    Jan 11 at 18:58
















1












1








1





$begingroup$


Seems like I do understand covariant the $Hom(A, -)$ and contravariant $Hom(-, A)$ functors. But when it comes down to the $Hom(-, -)$, then I am getting lost:



1) How does it work? What does it do with objects and how does it treat morphisms? [I assume it takes a product category $C times C$ as an input, but still underlaying details are unclear: I can't realize what is fixed and what is free.]



2) Why is it common to state that $Hom(-, -)$ is a kind of $C^{op} times C mapsto mathbb{SET}$ functor? [My assumption is that it because iterating over the first argument, having the second one fixed (i.e. $Hom(A, A'), Hom(B, A'), Hom(C, A'), ...$ naturally gives a contravariant functor, while same kind of iteration on the second argument - gives a covariant functor.]



2.1) Is it a matter of preference? In other words, is it safe to say that $C times C^{op} mapsto mathbb{SET}$ is equally valid definition? [I assume it is not matter of preference, but still better to ask.]



Could someone provide crystall clear definition of what $Hom(-, -)$ is and how should I think of it as an instance of $C times C^{op} mapsto mathbb{SET}$ functors?










share|cite|improve this question









$endgroup$




Seems like I do understand covariant the $Hom(A, -)$ and contravariant $Hom(-, A)$ functors. But when it comes down to the $Hom(-, -)$, then I am getting lost:



1) How does it work? What does it do with objects and how does it treat morphisms? [I assume it takes a product category $C times C$ as an input, but still underlaying details are unclear: I can't realize what is fixed and what is free.]



2) Why is it common to state that $Hom(-, -)$ is a kind of $C^{op} times C mapsto mathbb{SET}$ functor? [My assumption is that it because iterating over the first argument, having the second one fixed (i.e. $Hom(A, A'), Hom(B, A'), Hom(C, A'), ...$ naturally gives a contravariant functor, while same kind of iteration on the second argument - gives a covariant functor.]



2.1) Is it a matter of preference? In other words, is it safe to say that $C times C^{op} mapsto mathbb{SET}$ is equally valid definition? [I assume it is not matter of preference, but still better to ask.]



Could someone provide crystall clear definition of what $Hom(-, -)$ is and how should I think of it as an instance of $C times C^{op} mapsto mathbb{SET}$ functors?







category-theory definition






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asked Jan 11 at 18:46









Sereja BogolubovSereja Bogolubov

610211




610211








  • 1




    $begingroup$
    Re: (2) it's for the reason you say. Given a pair of maps $f:Ato B$ and $g:Cto D$, $(f,g)$ is a morphism $(B,C)to(A,D)$ in $C^{op}times C$; in $C$ composition $gcirc-circ f$ gives you a function $hom(B,C)tohom(A,D)$.
    $endgroup$
    – Malice Vidrine
    Jan 11 at 18:58
















  • 1




    $begingroup$
    Re: (2) it's for the reason you say. Given a pair of maps $f:Ato B$ and $g:Cto D$, $(f,g)$ is a morphism $(B,C)to(A,D)$ in $C^{op}times C$; in $C$ composition $gcirc-circ f$ gives you a function $hom(B,C)tohom(A,D)$.
    $endgroup$
    – Malice Vidrine
    Jan 11 at 18:58










1




1




$begingroup$
Re: (2) it's for the reason you say. Given a pair of maps $f:Ato B$ and $g:Cto D$, $(f,g)$ is a morphism $(B,C)to(A,D)$ in $C^{op}times C$; in $C$ composition $gcirc-circ f$ gives you a function $hom(B,C)tohom(A,D)$.
$endgroup$
– Malice Vidrine
Jan 11 at 18:58






$begingroup$
Re: (2) it's for the reason you say. Given a pair of maps $f:Ato B$ and $g:Cto D$, $(f,g)$ is a morphism $(B,C)to(A,D)$ in $C^{op}times C$; in $C$ composition $gcirc-circ f$ gives you a function $hom(B,C)tohom(A,D)$.
$endgroup$
– Malice Vidrine
Jan 11 at 18:58












1 Answer
1






active

oldest

votes


















3












$begingroup$

Indeed, a bifunctor is simply a functor $F$ defined on a product category. However, $Hom$ is not a functor, when defined on $mathcal{C} times mathcal{C}$: it sends a morphism $(A,C)to(B,C)$ to a morphism $Hom(B,C)to Hom(A,C)$, so it can't be a covariant functor (the arrow is in the wrong direction), but it also can't be a contravariant functor, for essentially the same reason: it sends $(A,B)to(A,C)$ to a morphism $Hom(A,B)to Hom(A,C)$. To fix this, we need to make either both factors covariant, or both factors contravariant. The way to do this, in either case, is to define one of the factors on the opposite category. Because people like covariant functors more than contravariant functors, they tend to choose the first, to make $Hom$ covariant. It would, indeed, be perfectly valid to define $Hom$ to be a contravariant functor from $mathcal{C} times mathcal{C}^{op}$, it's just a bit odd.



As for $Hom$ specifically: it is the functor from $mathcal{C}^{op}times mathcal{C}$ to $mathbb{SET}$, such that for every $(A,B)$ in $C^{op} times C$ (so $A$ and $B$ are both objects of $C$ (hence of $C^{op}$)), $Hom(A,B)$ is the set of all morphisms from $A$ to $B$, and for every morphism $(f,g): (A,B)to(C,D)$ of $C^{op}times C$ (so $f: C to A$, $g: B to D$ in $mathcal{C}$), $Hom(f,g)$ is the morphism of sets $Hom(A,B) to Hom(C,D)$ sending a morphism (of $mathcal{C}$) $varphi: A to B$ to the morphism $gcircvarphicirc fin Hom(C,D)$.



Alternatively, treating $Hom$ as a contravariant functor $mathcal{C}timesmathcal{C}^{op} to mathbf{SET}$: it is the functor from $C times C^{op}$ to $mathbb{SET}$, such that for every $(A,B)$ in $C times C^{op}$ (so $A$ and $B$ are both objects of $C$ (hence of $C^{op}$)), $Hom(A,B)$ is the set of all morphisms from $A$ to $B$, and for every morphism $(f,g): (A,B)to(C,D)$ of $C times C^{op}$ (so $f: A to C$, $g: D to B$ in $mathcal{C}$), $Hom(f,g)$ is the morphism of sets $Hom(C,D)to Hom(A,B)$ sending a morphism (of $mathcal{C}$) $psi: C to D$ to the morphism $gcircpsicirc fin Hom(A,B)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for answer, but I still can't get #1 from your explanation: does $Hom(-, -)$ imply that either it's left argument, or right has to be fixed, while the other is allowed to be iterated? Given a pair of objects $(A, B) in mathbb{C}^{op} times mathbb{C}$, what would $Hom(-, -)(A, B)$ do with it?
    $endgroup$
    – Sereja Bogolubov
    Jan 12 at 9:34












  • $begingroup$
    @SerejaBogolubov Not at all. It's just a functor, like any other functor, that happens to be defined on a product category. "Iterating" the argument is just evaluating it at different points, and you can evaluate it at any points you like.
    $endgroup$
    – user3482749
    Jan 12 at 11:28










  • $begingroup$
    So, how would it map $(A, B) rightarrow (A', B')$ then? Could you please provide a concrete "toy" example?
    $endgroup$
    – Sereja Bogolubov
    Jan 12 at 17:58












  • $begingroup$
    As I say, that morphism is really a pair of morphisms $f: A to A'$ and $g: B to B'$, which is mapped to the morphism (of sets) from $Hom(A,B)$ to $Hom(A',B')$ that takes a morphism $varphi: A to B$ and sends it to $gcircvarphicirc f$ (assuming you pick the usual covariant version, with the "op" in the first factor).
    $endgroup$
    – user3482749
    Jan 12 at 19:47










  • $begingroup$
    For a toy example, work in the category of abelian groups, take $A=B=mathbb{Z}^2$, $C=D= mathbb{Z}$, and our morphism to be the diagonal injection of $C = mathbb{Z}$ into $A = mathbb{Z}^2$ in the first position, and the projection onto the first coordinate in the second position, Then for any homomorphism $varphi: A to B$, $Hom(varphi)$ is the morphism from $C$ to $D$ that takes some $c in C = mathbb{Z}$, duplicates it to $(c,c) in A = mathbb{Z}^2$, maps it by $varphi$ to $varphi(c,c) in B = mathbb{Z}^2$, then projects that to its first coordinate, which is in $D = mathbb{Z}$.
    $endgroup$
    – user3482749
    Jan 12 at 19:47











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Indeed, a bifunctor is simply a functor $F$ defined on a product category. However, $Hom$ is not a functor, when defined on $mathcal{C} times mathcal{C}$: it sends a morphism $(A,C)to(B,C)$ to a morphism $Hom(B,C)to Hom(A,C)$, so it can't be a covariant functor (the arrow is in the wrong direction), but it also can't be a contravariant functor, for essentially the same reason: it sends $(A,B)to(A,C)$ to a morphism $Hom(A,B)to Hom(A,C)$. To fix this, we need to make either both factors covariant, or both factors contravariant. The way to do this, in either case, is to define one of the factors on the opposite category. Because people like covariant functors more than contravariant functors, they tend to choose the first, to make $Hom$ covariant. It would, indeed, be perfectly valid to define $Hom$ to be a contravariant functor from $mathcal{C} times mathcal{C}^{op}$, it's just a bit odd.



As for $Hom$ specifically: it is the functor from $mathcal{C}^{op}times mathcal{C}$ to $mathbb{SET}$, such that for every $(A,B)$ in $C^{op} times C$ (so $A$ and $B$ are both objects of $C$ (hence of $C^{op}$)), $Hom(A,B)$ is the set of all morphisms from $A$ to $B$, and for every morphism $(f,g): (A,B)to(C,D)$ of $C^{op}times C$ (so $f: C to A$, $g: B to D$ in $mathcal{C}$), $Hom(f,g)$ is the morphism of sets $Hom(A,B) to Hom(C,D)$ sending a morphism (of $mathcal{C}$) $varphi: A to B$ to the morphism $gcircvarphicirc fin Hom(C,D)$.



Alternatively, treating $Hom$ as a contravariant functor $mathcal{C}timesmathcal{C}^{op} to mathbf{SET}$: it is the functor from $C times C^{op}$ to $mathbb{SET}$, such that for every $(A,B)$ in $C times C^{op}$ (so $A$ and $B$ are both objects of $C$ (hence of $C^{op}$)), $Hom(A,B)$ is the set of all morphisms from $A$ to $B$, and for every morphism $(f,g): (A,B)to(C,D)$ of $C times C^{op}$ (so $f: A to C$, $g: D to B$ in $mathcal{C}$), $Hom(f,g)$ is the morphism of sets $Hom(C,D)to Hom(A,B)$ sending a morphism (of $mathcal{C}$) $psi: C to D$ to the morphism $gcircpsicirc fin Hom(A,B)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for answer, but I still can't get #1 from your explanation: does $Hom(-, -)$ imply that either it's left argument, or right has to be fixed, while the other is allowed to be iterated? Given a pair of objects $(A, B) in mathbb{C}^{op} times mathbb{C}$, what would $Hom(-, -)(A, B)$ do with it?
    $endgroup$
    – Sereja Bogolubov
    Jan 12 at 9:34












  • $begingroup$
    @SerejaBogolubov Not at all. It's just a functor, like any other functor, that happens to be defined on a product category. "Iterating" the argument is just evaluating it at different points, and you can evaluate it at any points you like.
    $endgroup$
    – user3482749
    Jan 12 at 11:28










  • $begingroup$
    So, how would it map $(A, B) rightarrow (A', B')$ then? Could you please provide a concrete "toy" example?
    $endgroup$
    – Sereja Bogolubov
    Jan 12 at 17:58












  • $begingroup$
    As I say, that morphism is really a pair of morphisms $f: A to A'$ and $g: B to B'$, which is mapped to the morphism (of sets) from $Hom(A,B)$ to $Hom(A',B')$ that takes a morphism $varphi: A to B$ and sends it to $gcircvarphicirc f$ (assuming you pick the usual covariant version, with the "op" in the first factor).
    $endgroup$
    – user3482749
    Jan 12 at 19:47










  • $begingroup$
    For a toy example, work in the category of abelian groups, take $A=B=mathbb{Z}^2$, $C=D= mathbb{Z}$, and our morphism to be the diagonal injection of $C = mathbb{Z}$ into $A = mathbb{Z}^2$ in the first position, and the projection onto the first coordinate in the second position, Then for any homomorphism $varphi: A to B$, $Hom(varphi)$ is the morphism from $C$ to $D$ that takes some $c in C = mathbb{Z}$, duplicates it to $(c,c) in A = mathbb{Z}^2$, maps it by $varphi$ to $varphi(c,c) in B = mathbb{Z}^2$, then projects that to its first coordinate, which is in $D = mathbb{Z}$.
    $endgroup$
    – user3482749
    Jan 12 at 19:47
















3












$begingroup$

Indeed, a bifunctor is simply a functor $F$ defined on a product category. However, $Hom$ is not a functor, when defined on $mathcal{C} times mathcal{C}$: it sends a morphism $(A,C)to(B,C)$ to a morphism $Hom(B,C)to Hom(A,C)$, so it can't be a covariant functor (the arrow is in the wrong direction), but it also can't be a contravariant functor, for essentially the same reason: it sends $(A,B)to(A,C)$ to a morphism $Hom(A,B)to Hom(A,C)$. To fix this, we need to make either both factors covariant, or both factors contravariant. The way to do this, in either case, is to define one of the factors on the opposite category. Because people like covariant functors more than contravariant functors, they tend to choose the first, to make $Hom$ covariant. It would, indeed, be perfectly valid to define $Hom$ to be a contravariant functor from $mathcal{C} times mathcal{C}^{op}$, it's just a bit odd.



As for $Hom$ specifically: it is the functor from $mathcal{C}^{op}times mathcal{C}$ to $mathbb{SET}$, such that for every $(A,B)$ in $C^{op} times C$ (so $A$ and $B$ are both objects of $C$ (hence of $C^{op}$)), $Hom(A,B)$ is the set of all morphisms from $A$ to $B$, and for every morphism $(f,g): (A,B)to(C,D)$ of $C^{op}times C$ (so $f: C to A$, $g: B to D$ in $mathcal{C}$), $Hom(f,g)$ is the morphism of sets $Hom(A,B) to Hom(C,D)$ sending a morphism (of $mathcal{C}$) $varphi: A to B$ to the morphism $gcircvarphicirc fin Hom(C,D)$.



Alternatively, treating $Hom$ as a contravariant functor $mathcal{C}timesmathcal{C}^{op} to mathbf{SET}$: it is the functor from $C times C^{op}$ to $mathbb{SET}$, such that for every $(A,B)$ in $C times C^{op}$ (so $A$ and $B$ are both objects of $C$ (hence of $C^{op}$)), $Hom(A,B)$ is the set of all morphisms from $A$ to $B$, and for every morphism $(f,g): (A,B)to(C,D)$ of $C times C^{op}$ (so $f: A to C$, $g: D to B$ in $mathcal{C}$), $Hom(f,g)$ is the morphism of sets $Hom(C,D)to Hom(A,B)$ sending a morphism (of $mathcal{C}$) $psi: C to D$ to the morphism $gcircpsicirc fin Hom(A,B)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for answer, but I still can't get #1 from your explanation: does $Hom(-, -)$ imply that either it's left argument, or right has to be fixed, while the other is allowed to be iterated? Given a pair of objects $(A, B) in mathbb{C}^{op} times mathbb{C}$, what would $Hom(-, -)(A, B)$ do with it?
    $endgroup$
    – Sereja Bogolubov
    Jan 12 at 9:34












  • $begingroup$
    @SerejaBogolubov Not at all. It's just a functor, like any other functor, that happens to be defined on a product category. "Iterating" the argument is just evaluating it at different points, and you can evaluate it at any points you like.
    $endgroup$
    – user3482749
    Jan 12 at 11:28










  • $begingroup$
    So, how would it map $(A, B) rightarrow (A', B')$ then? Could you please provide a concrete "toy" example?
    $endgroup$
    – Sereja Bogolubov
    Jan 12 at 17:58












  • $begingroup$
    As I say, that morphism is really a pair of morphisms $f: A to A'$ and $g: B to B'$, which is mapped to the morphism (of sets) from $Hom(A,B)$ to $Hom(A',B')$ that takes a morphism $varphi: A to B$ and sends it to $gcircvarphicirc f$ (assuming you pick the usual covariant version, with the "op" in the first factor).
    $endgroup$
    – user3482749
    Jan 12 at 19:47










  • $begingroup$
    For a toy example, work in the category of abelian groups, take $A=B=mathbb{Z}^2$, $C=D= mathbb{Z}$, and our morphism to be the diagonal injection of $C = mathbb{Z}$ into $A = mathbb{Z}^2$ in the first position, and the projection onto the first coordinate in the second position, Then for any homomorphism $varphi: A to B$, $Hom(varphi)$ is the morphism from $C$ to $D$ that takes some $c in C = mathbb{Z}$, duplicates it to $(c,c) in A = mathbb{Z}^2$, maps it by $varphi$ to $varphi(c,c) in B = mathbb{Z}^2$, then projects that to its first coordinate, which is in $D = mathbb{Z}$.
    $endgroup$
    – user3482749
    Jan 12 at 19:47














3












3








3





$begingroup$

Indeed, a bifunctor is simply a functor $F$ defined on a product category. However, $Hom$ is not a functor, when defined on $mathcal{C} times mathcal{C}$: it sends a morphism $(A,C)to(B,C)$ to a morphism $Hom(B,C)to Hom(A,C)$, so it can't be a covariant functor (the arrow is in the wrong direction), but it also can't be a contravariant functor, for essentially the same reason: it sends $(A,B)to(A,C)$ to a morphism $Hom(A,B)to Hom(A,C)$. To fix this, we need to make either both factors covariant, or both factors contravariant. The way to do this, in either case, is to define one of the factors on the opposite category. Because people like covariant functors more than contravariant functors, they tend to choose the first, to make $Hom$ covariant. It would, indeed, be perfectly valid to define $Hom$ to be a contravariant functor from $mathcal{C} times mathcal{C}^{op}$, it's just a bit odd.



As for $Hom$ specifically: it is the functor from $mathcal{C}^{op}times mathcal{C}$ to $mathbb{SET}$, such that for every $(A,B)$ in $C^{op} times C$ (so $A$ and $B$ are both objects of $C$ (hence of $C^{op}$)), $Hom(A,B)$ is the set of all morphisms from $A$ to $B$, and for every morphism $(f,g): (A,B)to(C,D)$ of $C^{op}times C$ (so $f: C to A$, $g: B to D$ in $mathcal{C}$), $Hom(f,g)$ is the morphism of sets $Hom(A,B) to Hom(C,D)$ sending a morphism (of $mathcal{C}$) $varphi: A to B$ to the morphism $gcircvarphicirc fin Hom(C,D)$.



Alternatively, treating $Hom$ as a contravariant functor $mathcal{C}timesmathcal{C}^{op} to mathbf{SET}$: it is the functor from $C times C^{op}$ to $mathbb{SET}$, such that for every $(A,B)$ in $C times C^{op}$ (so $A$ and $B$ are both objects of $C$ (hence of $C^{op}$)), $Hom(A,B)$ is the set of all morphisms from $A$ to $B$, and for every morphism $(f,g): (A,B)to(C,D)$ of $C times C^{op}$ (so $f: A to C$, $g: D to B$ in $mathcal{C}$), $Hom(f,g)$ is the morphism of sets $Hom(C,D)to Hom(A,B)$ sending a morphism (of $mathcal{C}$) $psi: C to D$ to the morphism $gcircpsicirc fin Hom(A,B)$.






share|cite|improve this answer









$endgroup$



Indeed, a bifunctor is simply a functor $F$ defined on a product category. However, $Hom$ is not a functor, when defined on $mathcal{C} times mathcal{C}$: it sends a morphism $(A,C)to(B,C)$ to a morphism $Hom(B,C)to Hom(A,C)$, so it can't be a covariant functor (the arrow is in the wrong direction), but it also can't be a contravariant functor, for essentially the same reason: it sends $(A,B)to(A,C)$ to a morphism $Hom(A,B)to Hom(A,C)$. To fix this, we need to make either both factors covariant, or both factors contravariant. The way to do this, in either case, is to define one of the factors on the opposite category. Because people like covariant functors more than contravariant functors, they tend to choose the first, to make $Hom$ covariant. It would, indeed, be perfectly valid to define $Hom$ to be a contravariant functor from $mathcal{C} times mathcal{C}^{op}$, it's just a bit odd.



As for $Hom$ specifically: it is the functor from $mathcal{C}^{op}times mathcal{C}$ to $mathbb{SET}$, such that for every $(A,B)$ in $C^{op} times C$ (so $A$ and $B$ are both objects of $C$ (hence of $C^{op}$)), $Hom(A,B)$ is the set of all morphisms from $A$ to $B$, and for every morphism $(f,g): (A,B)to(C,D)$ of $C^{op}times C$ (so $f: C to A$, $g: B to D$ in $mathcal{C}$), $Hom(f,g)$ is the morphism of sets $Hom(A,B) to Hom(C,D)$ sending a morphism (of $mathcal{C}$) $varphi: A to B$ to the morphism $gcircvarphicirc fin Hom(C,D)$.



Alternatively, treating $Hom$ as a contravariant functor $mathcal{C}timesmathcal{C}^{op} to mathbf{SET}$: it is the functor from $C times C^{op}$ to $mathbb{SET}$, such that for every $(A,B)$ in $C times C^{op}$ (so $A$ and $B$ are both objects of $C$ (hence of $C^{op}$)), $Hom(A,B)$ is the set of all morphisms from $A$ to $B$, and for every morphism $(f,g): (A,B)to(C,D)$ of $C times C^{op}$ (so $f: A to C$, $g: D to B$ in $mathcal{C}$), $Hom(f,g)$ is the morphism of sets $Hom(C,D)to Hom(A,B)$ sending a morphism (of $mathcal{C}$) $psi: C to D$ to the morphism $gcircpsicirc fin Hom(A,B)$.







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answered Jan 11 at 19:11









user3482749user3482749

4,296919




4,296919












  • $begingroup$
    Thanks for answer, but I still can't get #1 from your explanation: does $Hom(-, -)$ imply that either it's left argument, or right has to be fixed, while the other is allowed to be iterated? Given a pair of objects $(A, B) in mathbb{C}^{op} times mathbb{C}$, what would $Hom(-, -)(A, B)$ do with it?
    $endgroup$
    – Sereja Bogolubov
    Jan 12 at 9:34












  • $begingroup$
    @SerejaBogolubov Not at all. It's just a functor, like any other functor, that happens to be defined on a product category. "Iterating" the argument is just evaluating it at different points, and you can evaluate it at any points you like.
    $endgroup$
    – user3482749
    Jan 12 at 11:28










  • $begingroup$
    So, how would it map $(A, B) rightarrow (A', B')$ then? Could you please provide a concrete "toy" example?
    $endgroup$
    – Sereja Bogolubov
    Jan 12 at 17:58












  • $begingroup$
    As I say, that morphism is really a pair of morphisms $f: A to A'$ and $g: B to B'$, which is mapped to the morphism (of sets) from $Hom(A,B)$ to $Hom(A',B')$ that takes a morphism $varphi: A to B$ and sends it to $gcircvarphicirc f$ (assuming you pick the usual covariant version, with the "op" in the first factor).
    $endgroup$
    – user3482749
    Jan 12 at 19:47










  • $begingroup$
    For a toy example, work in the category of abelian groups, take $A=B=mathbb{Z}^2$, $C=D= mathbb{Z}$, and our morphism to be the diagonal injection of $C = mathbb{Z}$ into $A = mathbb{Z}^2$ in the first position, and the projection onto the first coordinate in the second position, Then for any homomorphism $varphi: A to B$, $Hom(varphi)$ is the morphism from $C$ to $D$ that takes some $c in C = mathbb{Z}$, duplicates it to $(c,c) in A = mathbb{Z}^2$, maps it by $varphi$ to $varphi(c,c) in B = mathbb{Z}^2$, then projects that to its first coordinate, which is in $D = mathbb{Z}$.
    $endgroup$
    – user3482749
    Jan 12 at 19:47


















  • $begingroup$
    Thanks for answer, but I still can't get #1 from your explanation: does $Hom(-, -)$ imply that either it's left argument, or right has to be fixed, while the other is allowed to be iterated? Given a pair of objects $(A, B) in mathbb{C}^{op} times mathbb{C}$, what would $Hom(-, -)(A, B)$ do with it?
    $endgroup$
    – Sereja Bogolubov
    Jan 12 at 9:34












  • $begingroup$
    @SerejaBogolubov Not at all. It's just a functor, like any other functor, that happens to be defined on a product category. "Iterating" the argument is just evaluating it at different points, and you can evaluate it at any points you like.
    $endgroup$
    – user3482749
    Jan 12 at 11:28










  • $begingroup$
    So, how would it map $(A, B) rightarrow (A', B')$ then? Could you please provide a concrete "toy" example?
    $endgroup$
    – Sereja Bogolubov
    Jan 12 at 17:58












  • $begingroup$
    As I say, that morphism is really a pair of morphisms $f: A to A'$ and $g: B to B'$, which is mapped to the morphism (of sets) from $Hom(A,B)$ to $Hom(A',B')$ that takes a morphism $varphi: A to B$ and sends it to $gcircvarphicirc f$ (assuming you pick the usual covariant version, with the "op" in the first factor).
    $endgroup$
    – user3482749
    Jan 12 at 19:47










  • $begingroup$
    For a toy example, work in the category of abelian groups, take $A=B=mathbb{Z}^2$, $C=D= mathbb{Z}$, and our morphism to be the diagonal injection of $C = mathbb{Z}$ into $A = mathbb{Z}^2$ in the first position, and the projection onto the first coordinate in the second position, Then for any homomorphism $varphi: A to B$, $Hom(varphi)$ is the morphism from $C$ to $D$ that takes some $c in C = mathbb{Z}$, duplicates it to $(c,c) in A = mathbb{Z}^2$, maps it by $varphi$ to $varphi(c,c) in B = mathbb{Z}^2$, then projects that to its first coordinate, which is in $D = mathbb{Z}$.
    $endgroup$
    – user3482749
    Jan 12 at 19:47
















$begingroup$
Thanks for answer, but I still can't get #1 from your explanation: does $Hom(-, -)$ imply that either it's left argument, or right has to be fixed, while the other is allowed to be iterated? Given a pair of objects $(A, B) in mathbb{C}^{op} times mathbb{C}$, what would $Hom(-, -)(A, B)$ do with it?
$endgroup$
– Sereja Bogolubov
Jan 12 at 9:34






$begingroup$
Thanks for answer, but I still can't get #1 from your explanation: does $Hom(-, -)$ imply that either it's left argument, or right has to be fixed, while the other is allowed to be iterated? Given a pair of objects $(A, B) in mathbb{C}^{op} times mathbb{C}$, what would $Hom(-, -)(A, B)$ do with it?
$endgroup$
– Sereja Bogolubov
Jan 12 at 9:34














$begingroup$
@SerejaBogolubov Not at all. It's just a functor, like any other functor, that happens to be defined on a product category. "Iterating" the argument is just evaluating it at different points, and you can evaluate it at any points you like.
$endgroup$
– user3482749
Jan 12 at 11:28




$begingroup$
@SerejaBogolubov Not at all. It's just a functor, like any other functor, that happens to be defined on a product category. "Iterating" the argument is just evaluating it at different points, and you can evaluate it at any points you like.
$endgroup$
– user3482749
Jan 12 at 11:28












$begingroup$
So, how would it map $(A, B) rightarrow (A', B')$ then? Could you please provide a concrete "toy" example?
$endgroup$
– Sereja Bogolubov
Jan 12 at 17:58






$begingroup$
So, how would it map $(A, B) rightarrow (A', B')$ then? Could you please provide a concrete "toy" example?
$endgroup$
– Sereja Bogolubov
Jan 12 at 17:58














$begingroup$
As I say, that morphism is really a pair of morphisms $f: A to A'$ and $g: B to B'$, which is mapped to the morphism (of sets) from $Hom(A,B)$ to $Hom(A',B')$ that takes a morphism $varphi: A to B$ and sends it to $gcircvarphicirc f$ (assuming you pick the usual covariant version, with the "op" in the first factor).
$endgroup$
– user3482749
Jan 12 at 19:47




$begingroup$
As I say, that morphism is really a pair of morphisms $f: A to A'$ and $g: B to B'$, which is mapped to the morphism (of sets) from $Hom(A,B)$ to $Hom(A',B')$ that takes a morphism $varphi: A to B$ and sends it to $gcircvarphicirc f$ (assuming you pick the usual covariant version, with the "op" in the first factor).
$endgroup$
– user3482749
Jan 12 at 19:47












$begingroup$
For a toy example, work in the category of abelian groups, take $A=B=mathbb{Z}^2$, $C=D= mathbb{Z}$, and our morphism to be the diagonal injection of $C = mathbb{Z}$ into $A = mathbb{Z}^2$ in the first position, and the projection onto the first coordinate in the second position, Then for any homomorphism $varphi: A to B$, $Hom(varphi)$ is the morphism from $C$ to $D$ that takes some $c in C = mathbb{Z}$, duplicates it to $(c,c) in A = mathbb{Z}^2$, maps it by $varphi$ to $varphi(c,c) in B = mathbb{Z}^2$, then projects that to its first coordinate, which is in $D = mathbb{Z}$.
$endgroup$
– user3482749
Jan 12 at 19:47




$begingroup$
For a toy example, work in the category of abelian groups, take $A=B=mathbb{Z}^2$, $C=D= mathbb{Z}$, and our morphism to be the diagonal injection of $C = mathbb{Z}$ into $A = mathbb{Z}^2$ in the first position, and the projection onto the first coordinate in the second position, Then for any homomorphism $varphi: A to B$, $Hom(varphi)$ is the morphism from $C$ to $D$ that takes some $c in C = mathbb{Z}$, duplicates it to $(c,c) in A = mathbb{Z}^2$, maps it by $varphi$ to $varphi(c,c) in B = mathbb{Z}^2$, then projects that to its first coordinate, which is in $D = mathbb{Z}$.
$endgroup$
– user3482749
Jan 12 at 19:47


















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