Rotman's Homological Algebra Text, Def of Adjoint Functors

Question

I've been going thru Rotman's text Homological Algebra and one detail of his proof bothers me. The proof is

The thing that bothers me is that in his definition of adjoint pair before this

the downward maps are only assumed to be a bijection of Hom-set, but he claims that it is an isomorphism in the proof?? Did he mean to say that the map is an isomorphism, not just bijection, in the definition of adjoint pair instead?

Thank you all.

Answer 1

In the category of modules over a ring $R$ (in fact, in an abelian category in general), addition in $\hom(A,B)$ can be defined as follows :

$\hom(A,B)\times \hom(A,B)\to \hom(A\oplus A, B) \to \hom(A,B)$

where the first map is obtained by the universal property of the direct sum, and the second map is obtained by the diagonal $A\to A\times A = A\oplus A$ (this can be alternatively described as the sum of the two inclusions $A\to A\oplus A$).

In $_R\mathfrak M$ you can check that by hand, and in an abelian category you can simply use the alternative description and the bilinearity of composition; but you have to check that the two descriptions I gave actually coincide.

The main point being that this description of addition in $\hom(A,B)$ is "categorical". In fact this suffices to check that $\hom(F(A),B)\to \hom(A,G(B))$ is a group morphism whenever $F\dashv G$ is an adunction.

Indeed we want to check that

$\require{AMScd}\begin{CD} \hom(F(A),B)\times \hom(F(A),B) @>>> \hom(A,G(B))\times \hom(A,G(B)) \\ @VVV @VVV \\ \hom(F(A),B) @>>> \hom(A,G(B))\end{CD}$

commutes, so we decompose it as follows

$\require{AMScd}\begin{CD} \hom(F(A),B)\times \hom(F(A),B) @>>> \hom(A,G(B))\times \hom(A,G(B)) \\ @VVV @VVV \\ \hom(F(A)\oplus F(A), B) @>>> \hom(A\oplus A, G(B)) \\ @VVV @VVV \\ \hom(F(A),B) @>>> \hom(A,G(B))\end{CD}$

and we treat both squares independently :

• note that in the top squares, since the vertical arrows are obtained by universal properties, they are isomorphisms, so we may turn them around and check the new commutativity :

$\require{AMScd}\begin{CD} \hom(F(A),B)\times \hom(F(A),B) @>>> \hom(A,G(B))\times \hom(A,G(B)) \\ @AAA @AAA \\ \hom(F(A)\oplus F(A), B) @>>> \hom(A\oplus A, G(B))\end{CD}$

now we want to check that two maps into a product are the same, so we can treat each coordinate separately :

$\require{AMScd}\begin{CD} \hom(F(A),B) @>>> \hom(A,G(B)) \\ @AAA @AAA \\ \hom(F(A)\oplus F(A), B) @>>> \hom(A\oplus A, G(B))\end{CD}$

Now note that $F$ commutes with direct sums, which means we can choose $F(A\oplus A)$ as a model for $F(A)\oplus F(A)$, with the inclusions $F(A)\to F(A\oplus A)$ being given by $F$ of the inclusions $A\to A\oplus A$. It follows that for the top square we only need to show that

$\require{AMScd}\begin{CD} \hom(F(A),B) @>>> \hom(A,G(B)) \\ @AAA @AAA \\ \hom(F(A\oplus A), B) @>>> \hom(A\oplus A, G(B))\end{CD}$

commutes. But now this is true because the adjunction isomorphism $\hom(F(X),Y)\cong \hom(X,G(Y))$ is natural in $X$, and this square is nothing but a naturality square.

So the top square commutes.

• The bottom square is treated in exactly the same way : consider $F(A\oplus A)$ as a model for $F(A)\oplus F(A)$, and prove that the diagonal $F(A)\to F(A\oplus A)$ is just $F$ of the diagonal $A\to A\oplus A$ (this follows from the fact that $F$ preserves finite products, because they are just finite coproducts in $_R\mathfrak{M}$)

So both squares commute, so the whole outer rectangle commutes, and this is precisely the definition of being a group morphism