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What really is the difference between a Homomorphism between categories and a Functor? I understand that they are both structure preserving mapping of the objects and arrows between the categories or am i missing something ?

vinothkr
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    What is your definition of a "homomorphism of categories"? – Zev Chonoles May 29 '13 at 06:35
  • I mean there exists a homomorphism between categories M and N then i dont see why i cant call that a Functor – vinothkr May 29 '13 at 06:39
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    You didn't answer my question - what is your definition of the term "homomorphism of categories"? Given two categories $M$ and $N$, what are the properties something must have to be a homomorphism from $M$ to $N$? – Zev Chonoles May 29 '13 at 06:40
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    Given h is a homomorphism h(a.b) = h(a).h(b) where a and b belongs to M and h(a) and h(b) belongs to N then Its a homomorphism – vinothkr May 29 '13 at 06:45
  • What does "$a.b$" mean? Are $a$ and $b$ objects or morphisms? – Zev Chonoles May 29 '13 at 06:47
  • a and b are morphisms. and a.b i mean is a composite. – vinothkr May 29 '13 at 06:48
  • then type a between dollar symbols, b between dollar symbols, and a \circ b between dollar symbols. – Ittay Weiss May 29 '13 at 07:01
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    Out of curiosity, where did you get this definition of a homomorphism between categories? – Ink May 29 '13 at 07:05
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    @vinothkr: Your "definition" of a homomorphism of categories is nonsense. When you ignore the identities, this is not the correct notion of a homomorphism. – Martin Brandenburg May 29 '13 at 10:40
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    if $h$($e$ $\circ$ $b$) = $h$($b$) = $h$($e$) $\circ$ $h$($b$) where $e$ is an identity in $M$ and $h$($e$) is also an identity in $N$ which implies the identity is preserved right? I thought thats just inferred – vinothkr May 29 '13 at 11:51
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    It's not. Read the answer by Zev Chonoles. – Martin Brandenburg May 29 '13 at 19:57
  • According to Zev Chonoles preserving identity is not required for Homomorphism. You should read that too i guess – vinothkr May 30 '13 at 06:19
  • @vinothkr: I am assuming that a homomorphism is not required to preserve identity morphisms; since I've never heard of the notion "homomorphism of categories" before, I am working from the definition you gave me. Martin is correct: my answer shows that just assuming that a map preserves compositions does not imply that it will preserve identity morphisms. – Zev Chonoles May 30 '13 at 06:22
  • I dont understand the part $c$ $\circ$ $c$ = $c$ but $c$ is not an identity. – vinothkr May 30 '13 at 06:28
  • @vinothkr: When we are making up a category, we can define the composition law any way we like as long as the identity morphisms work as they are supposed to, and associativity holds. The only morphisms in the category $N$ are $\mathrm{id}_z:z\to z$ and $c:z\to z$. You can check that all of the axioms a category is required to have are met. – Zev Chonoles May 30 '13 at 20:43

2 Answers2

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You are correct, functors are the structure preserving entities between categories. They could have been called homomorphisms. I don't know much about the reason for the terminology used, but I think that Mac Lane and Eilenberg, when inventing category theory, borrowed terminology from philosophy where category and functor are known (at least to some).

Polymorphism is common in mathematics. For instance, the structure preserving entities between topological spaces (though some may say these should actually be considered to be between frames) are called continuous rather than homomorphisms. And an invertible continuous function with continuous inverse is called a homeomorphism rather than an isomorphism. A structure preserving entity between vectors spaces is called a linear transformation. A structure preserving mappings between metric spaces is called a short map. And an invertible one is called an isometry rather than an isomorphism. Clearly, historical reasons play a role.

Specifically regarding categories, since categories don't have to be small, one can argue that a functor is not really a function. One can argue that a homomorphism must be a function that preserves structure, and thus one may claim that functors should not be called homomorphisms. In any case, the terminology, while not set in stone, is certainly well-established but that does not mean other possibilities do not make sense. Sometimes they make more sense.

Ittay Weiss
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    Sometimes functors are called morphisms of categories or even map of categories, and natural transformations are called morphisms of functors of even map of functors. – Martin Brandenburg May 29 '13 at 10:41
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If I understand your question and definitions correctly: no, a homomorphism of categories and a functor are not the same thing.

Your definition of "homomorphism of categories" seems to be: given categories $M$ and $N$, a homomorphism $h:M\to N$ is

  • a mapping from $\mathsf{ob}(M)$ to $\mathsf{ob}(N)$, together with
  • for every $x,y\in\mathsf{ob}(M)$, a mapping from $\mathrm{Hom}_M(x,y)$ to $\mathrm{Hom}_N(h(x),h(y))$ such that $$h(a\circ b)=h(a)\circ h(b)$$ for all $a\in \mathrm{Hom}_M(x,y)$ and $b\in \mathrm{Hom}_M(y,z)$.

A functor is then a homomorphism of categories that satisfies the additional requirement that

  • $h(\mathrm{id}_x)=\mathrm{id}_{h(x)}$ for all $x\in\mathsf{ob}(M)$.

Thus, being a functor is a strictly stronger property than being a homomorphism of categories. For example, let $M$ be the category with

  • $\mathsf{ob}(M)=\{x,y\}$
  • $\mathrm{Hom}_M(x,x)=\{\mathrm{id}_x\},\quad \mathrm{Hom}_M(x,y)=\{a\}, \quad \mathrm{Hom}_M(y,x)=\varnothing,\quad \mathrm{Hom}_M(y,y)=\{\mathrm{id}_y\}$

and let $N$ be the category with

  • $\mathsf{ob}(N)=\{z\}$
  • $\mathrm{Hom}_N(z,z)=\{\mathrm{id}_z,c\}$ where $c\circ c=c$.

Then the homomorphism of categories $h:M\to N$ defined by

  • $h(x)=z,\quad h(y)=z$
  • $h(\mathrm{id}_x)=c,\quad h(a)=c,\quad h(\mathrm{id}_y)=c$

is not a functor.

user1483891
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Zev Chonoles
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  • if the identity is taken to be the empty composition in the monoid of endo-arrows, then it's fine again. – Ittay Weiss May 29 '13 at 07:04
  • I don't think I understand your comment; a monoid comes with an identity element, and a monoid morphism must preserve identity elements. The identity element can be thought of as the empty composition, but this still does not prevent us from forming a semigroup morphism that is not a monoid morphism. – Zev Chonoles May 29 '13 at 07:12
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    I just mean that if you define a monoid homomorphism to preserve all compositions (including the empty one), then it (tautologically) preserves the unit. A function between monoids that preserves binary compositions but not (necessarily) the unit, is the same as a function that preserves all non-empty finite compositions. I'm not saying anything smart here, just that if one includes the empty composition as a composition, then composition preserving functions preserves units as well. – Ittay Weiss May 29 '13 at 07:42
  • Hm, so wikipedia seems to be mistaken about that: https://en.wikipedia.org/wiki/Homomorphism#Definition . Check out the last item in the list. – Vadim Samokhin Feb 29 '16 at 18:57
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    @Zapadlo: Wikipedia is correct. However, the OP is using a non-standard definition of "homomorphism of categories", see their comment below the question. What the OP is calling a homomorphism of categories is not a functor. – Zev Chonoles Feb 29 '16 at 19:49