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I am reading an example in Hungerford's Algebra (Section X.1).

If $A, B, C$ are left modules over a ring $R$, then the isomorphism of abelian groups $$\phi:\text{Hom}_{R}{(A\oplus B,C)}\cong\text{Hom}_{R}{(A,C)}\oplus \text{Hom}_{R}{(B,C)}$$ of Theorem IV.4.7 is natural. One may interpret the word "natural" here by fixing any two variables, say $A$ and $C$, and observing that for each module homomorphism $f:B\to B'$ the diagram $$\text{Hom}_{R}{(A\oplus B',C)} \stackrel{\phi}{\to} \text{Hom}_{R}{(A,C)}\oplus \text{Hom}_{R}{(B',C)}$$ $$\text{Hom}{(1_A\oplus f,1_C)} \downarrow~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \downarrow\text{Hom}{(1_A,1_C)}\oplus \text{Hom}{(f,1_C)}$$ $$\text{Hom}_{R}{(A\oplus B,C)} \stackrel{\phi}{\to} \text{Hom}_{R}{(A,C)}\oplus \text{Hom}_{R}{(B,C)}$$ is commutative, where $1_A\oplus f:A\oplus B\to A\oplus B'$ is given by $(a, b)\mapsto (a, f(b))$. Thus $\phi$ defines a natural isomorphism of the contravariant functors $S$ and $T$, where $$S(B)=\text{Hom}_{R}{(A\oplus B,C)}\text{ and }T(B)=\text{Hom}_{R}{(A,C)}\oplus \text{Hom}_{R}{(B,C)}.$$ One says that the isomorphism $\phi$ is natural in $B$. A similar argument shows that $\phi$ is natural in $A$ and $C$ as well.

My Question.

  • I think that $S(f)(g)(a, b)=(g\circ (1_A\oplus f))(a, b)$, where $a\in A, b\in B, g\in \text{Hom}_{R}{(A\oplus B',C)}$.
  • I don't understand what $\text{Hom}{(1_A\oplus f,1_C)}$ means.
  • $\text{Hom}{(1_A\oplus f,1_C)}(g)=g\circ (1_A\oplus f)$?
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2 Answers 2

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In general, if $\mathcal{C}$ is a category, then $\mathrm{Hom}_{\mathcal{C}}$ is a functor $\mathcal{C}^{\mathrm{op}} \times \mathcal{C} \to \mathbf{Set}$ (or $\mathbf{AbGp}$ or whatever enrichment you're working with).

The action of $\mathbf{Hom}_{\mathcal{C}}$ on morphisms is given by $$\mathrm{Hom}_{\mathcal{C}}(f : A \leftarrow B , g : C \to D) = g \circ (-) \circ f : \mathrm{Hom}_{\mathcal{C}}(A,C) \to \mathrm{Hom}_{\mathcal{C}}(B,D)$$ (Note the contravariance in the first argument.)

That is, for fixed $f : B \to A$ and $g : C \to D$, we have $$\mathrm{Hom}(f,g)(h) = g \circ h \circ f : B \to D$$ for all $h : A \to C$.

So in particular, you have $\mathrm{Hom}(1_A \oplus f, 1_C)(g) = 1_C \circ g \circ (1_A \oplus f) = g \circ (1_A \oplus f)$, as you suggest.

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  • $\begingroup$ Does $S(f)=\text{Hom}(1_A\oplus f, 1_C)$? $\endgroup$
    – bfhaha
    May 1, 2018 at 16:55
  • $\begingroup$ @bfhaha: Yes. $~$ $\endgroup$ May 1, 2018 at 17:13
  • $\begingroup$ Got it! Thanks. $\endgroup$
    – bfhaha
    May 1, 2018 at 17:15
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For a functor, like $\newcommand{\Hom}{\operatorname{Hom}}\Hom$ or $\newcommand{\dsum}{\oplus}\dsum$, it is common to use the functor to denote both the action on objects and maps. Thus if $f:A\to B$, $g:C\to D$, then $\Hom(f,g) : \Hom(B,C)\to \Hom(A,D)$ is the map $h\mapsto g\circ h \circ f$. Therefore, you are right in your guess.

This is entirely analagous to how $f\dsum g : A\dsum C \to B\dsum D$ is the map $(a,c)\mapsto (f(a),g(c))$, which is notation it appears you are familiar with.

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