Question on adjoint functors

Question

Can someone provide me an enlightenment on the following three statements? (I stumbled on them at the part dealing injective modules in a text of homological algebra.)

1) Let $F \dashv G \colon \mathcal{C} \to \mathcal{D}$ is a pair of adjoint functors and $F$ preserves monomorphisms. Then $G$ preserves injectives.

2) A left adjoint preserves epimorphisms.

3) A right adjoint preserves monomorphisms.

Seeing that the text gives no clues for the proofs, they must make an easy exercise. But I cannot make out how to show them.

Answer 1

General advice when dealing with this sort of exercise: Definitions, definitions, definitions. Always go back to the definitions. Write down explicitly what you want to prove, and in 95% of cases the proof will almost write itself down for you.

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For example for question #1: I will write down everything very formally. In a normal proof everything would be condensed in a few lines (cf. Hanno's answer), but hopefully this will help you in finding how to reason.

What is an injective? It's an object $I$ such that if for every morphism $f : A \to I$ and every monomorphism (notice how $F$ conveniently preserves monomorphisms?) $g : A \to B$, there exists a morphism $h : B \to I$ such that $h \circ g = f$ (write down the commutative diagram).

Now you want to show that if

• Data 1: $I \in \mathcal{D}$ is injective,

then $G(I) \in \mathcal{C}$ is injective. So by definition you take

• Data 2: any morphism $f : A \to G(I)$, and
• Data 3: any monomorphism $g : A \to B$,

and you look for some $h : B \to G(I)$ that makes the diagram commute.

Now you only have two hypotheses:

• Hyp 4: $F \dashv G$,
• Hyp 5: $F$ preserves monomorphisms.

There's only one monomorphism in this story, so apply your Hyp 5 to it:

• Fact 6: $F(g) : F(A) \to F(B)$ is a monomorphism.

You know that $I$ is injective. By Hyp 4, you've got a morphism $\tilde{f} : F(A) \to I$ naturally associated to $f$. By taking Fact 6 into account, and the Data 1 that $I$ is injective, you find that

• Fact 7: there exists a morphism $\phi : F(B) \to I$ such that $\phi \circ F(g) = \tilde{f}$.

Now you apply Hyp 4 again and you find a morphism $h = \bar\phi : B \to G(I)$. Since the adjunction $F \dashv G$ is natural, you finally find that $h \circ g = f$. Qed.

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Now if you follow the same pattern of reasoning you should find the answers to question 2 and 3.

Answer 2

Hint for (1): Note that an object $I\in{\mathscr D}$ is injective if and only if ${\mathscr D}(-,I): {\mathscr D}^{\text{op}}\to\textsf{Set}$ preserves epimorphisms (i.e., turns monomorphisms in ${\mathscr D}$ into surjective maps of sets) - and analogously for objects of ${\mathscr C}$. Now suppose $X\in{\mathscr D}$ is injective & that ${\mathscr F}$ preserves monomorphisms and check that by adjointness of ${\mathscr F}$ and ${\mathscr G}$ the functor ${\mathscr C}(-,{\mathscr G}X)$ also maps monomorphisms in ${\mathscr C}$ to epimorphisms.

Parts (2) and (3) are similar, noting that a morphism $f: X\to Y$ in ${\mathscr C}$ is, say, an epimorphism if and only if for all $Z\in{\mathscr C}$ the induced map of sets ${\mathscr C}(f,Z): {\mathscr C}(Y,Z)\to{\mathscr C}(X,Z)$ is injective, i.e. a monomorphism in $\textsf{Set}$. Similar for monomorphisms.

Answer 3

Although the previous answers contain hints how to prove 2) and 3), it seems desirable to give a complete proof.

We only prove 2), the proof of 3) is dual.

Let $F : \mathcal C \to \mathcal D$ be a left adjoint to some functor $G : \mathcal D \to \mathcal C$. This means that there is a natural bijection $$\tau_{X,Y} : \mathcal D(F(X),Y) \to \mathcal C(X,G(Y)), X \in ob(\mathcal C), Y \in ob(\mathcal D) . $$

Let $e : Z \to X$ be an epimorphism in $\mathcal C$. We have to show that $F(e) : F(Z) \to F(X)$ is an epimorphism in $\mathcal D$ which means that if $f_0, f_1 : F(X) \to Y$ are two morphisms in $\mathcal D$ such that $f_0 \circ F(e) = f_1 \circ F(e)$, then $f_0 = f_1$.

Consider the following commutative diagram:

$\require{AMScd}$ \begin{CD} \mathcal D(F(X),Y) @>{\tau_{X,Y}}>> \mathcal C(X,G(Y)) \\ @V{F(e)^*}VV @VV{e^*}V \\ \mathcal D(F(Z),Y) @>>{\tau_{Z,Y}}> \mathcal C(Z,G(Y)) \end{CD}

We are given $f_0, f_1 \in \mathcal D(F(X),Y)$ such that $F(e)^*(f_0) = f_0 \circ F(e) = f_1 \circ F(e) = F(e)^*(f_1)$. We conclude $$\tau_{X,Y}(f_0)\circ e = e^*(\tau_{Z,Y}(f_0)) = \tau_{Z,Y}(F(e)^*(f_0)) = \\ \tau_{Z,Y}(F(e)^*(f_1)) = e^*(\tau_{Z,Y}(f_1)) = \tau_{X,Y}(f_1)\circ e .$$ Thus $\tau_{X,Y}(f_0) = \tau_{X,Y}(f_1)$ because $e$ is an epimorphism. Hence $f_0 = f_1$ because $\tau_{X,Y}$ is a bijection.

A nice corollary is this:

A necessary condition for $F : \mathcal C \to \mathcal D$ having a right adjoint is that $F$ preserves epimorphisms.