Vtyjkyuk

The full question:

Suppose we have two categories $mathcal A$ and $mathcal B$ for which it makes sense to talk about injective and surjective maps (i.e., there is a faithful functor into the category $textbfSets$).

If $mathcal Fcolon mathcal Ato mathcal B$ is left adjoint to $mathcal Gcolonmathcal Btomathcal A$, so that
$$varphicolon operatornameHom_mathcal B(mathcal F-,-)oversetsimRightarrowoperatornameHom_mathcal A(-,mathcal G -),$$
then can we say that a map $fcolon mathcal FXto Y$ is injective (surjective) if and only if $varphi(f)$ is injective (surjective)?

For example, by Frobenius reciprocity, $operatornameInd^G_H-cong operatornameRes_H^G-$. Does this mean that a map
$$fcolon Bbb FGotimes Vto W $$
in $Bbb FG$-$operatornameMod$ is injective if and only if the corresponding map of $Bbb FH$-modules
$$tilde fcolon Vto operatornameRes^G_HW $$
is injective? Here, $tilde f(v)=f(1otimes v).$ This is a fact that I would love to be true but I'm having a hard time pulling apart.

Here are some counterexamples.

Let $F$ and $G$ be the free and forgetful functor for abelian groups.

Consider the map $x,y to G(mathbbZ)$ defined by sending $x$ and $y$ to $1$ and $-1$ respectively. This is injective.

Its transpose is a map $F(x,y) to mathbbZ$. But $F(x,y) cong mathbbZ oplus mathbbZ$, so this map clearly cannot be injective; e.g. the elements $0$ and $x+y$ are distinct but have the same image.

An even simpler counterexample is to start with $x to G(mathbbZ)$ sending $x$ to $0$.

To get a counterexample to surjectivity, consider any set $S$. The map $S to GF(S)$ is not surjective, but its transpose $F(S) to F(S)$ is.

The dual of this last example gives another counterexample to injectivity. For any abelian group, $G(A) to G(A)$ is injective, but $FG(A) to A$ is not.