In a follow-on comment, the answerer of this question opines that the standard notation used to represent $\hom$-functors is "bad". That's somewhat comforting, as it tells me my own confusion is caused by muddled notation, not lack of the necessary neurons on my part.
If you agree with this assessment, can you offer, simply for pedagogical purposes, an alternate notation that would be more consistent and descriptive that might trigger an a-ha moment for me?
UPDATE: stared at the above-linked SO question for a while, and back at the text I'm studying from (Abstract and Concrete Categories: the Joy of Cats, by Adamek, Herrlich & Strecker, p. 30). Here's the relevant section of the text:
For any category $\mathbf{A}$ and any $\mathbf{A}$-object $\mathbf{A}$, there is the covariant hom-functor $\hom(A, −) : \mathbf{A → Set}$, defined by $$\hom(A, −)(B \xrightarrow f C) = \hom(A, B) \xrightarrow {\hom(A,f)} \hom(A, C)$$ where $\hom(A, f)(g) = f \circ g.$
It looks to me like $$\hom(A,f)(g)$$ is intended as a rewrite of the left side of the equation above it $$\hom(A,-)(B \xrightarrow f C)$$ with $f$ substituted for the "-" and $g$ substituted for the $f$. But that's obviously wrong. To get the right sequence ($A, B, C$) of domains and codomains, $f$ in the final equation has to be the same as the $B \xrightarrow f C$ in the previous one, and $g$ has to be $A \xrightarrow g B$. But if that's so, why are the positions switched? (As I write this, the light is dawning about the meaning of the right side of the main equation, in which $g=\hom(A,B)$ is the argument passed to $\hom(A,f)$ and the result is the converted morphism $A \xrightarrow h C = \hom(A,C)$. But it still seems like the left side doesn't match the left side of the "where" clause equation.)
UPDATE 2: Thanks to the erudite responses below, my understanding is getting better. Here's my attempt to document the blanks I've filled in. It starts with a pretty picture:
- The objects in Set are in this case sets of A-morphisms originating at object $A$. So... the objects are sets of morphisms – a slightly disorienting fact initially to me and probably other Cat-Theory neophytes. I understood it from the beginning but I think it contributed to some head scratching.
- A co-variant hom-functor $\hom(A,-)$ can be thought of as generating new morphisms from $A$ to $C$, using existing $A \rightarrow B$ morphisms with a $B \rightarrow C$ “helper function” to complete the path. This is just good old composition, no more and no less, only viewed in a slightly weird way that brings the Set category into the mix.
- To transform any given morphism or object, we fill in the "–" in $\hom(A,–)$ with a specific object or morphism from A. That is, if we fill in with an object $B$, then $\hom(A,B)$ yields (in fact, is) the set of -- well, $\hom(A,B)$. If we fill in with A-morphism $f: A \rightarrow B$, then $\hom(A,-)$ converts A-morphism $f$ to a Set-morphism $\hom(A,f)$. Since the objects we're concerned with in Set are A-morphisms, the morphisms we care about in Set take us from one A-morphism to another A-morphism. This is accomplished via regular old composition: $\hom(A,f)$ gives us a Set-morphism that takes us from an A-morphism $g$ to $f \circ g$, which is of course another A-morphism.
- All that said, we haven't written down explicitly any function $F_o:$ Ob(A) $\rightarrow$ Ob(Set) to transform objects, or any $F_m:$ Mor(A) $\rightarrow$ Mor(Set) to transform morphisms. Since we're looking at functors, not functions, maybe this is what I should expect.