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 $\textbf{Sets}$).
If $\mathcal F\colon \mathcal A\to \mathcal B$ is left adjoint to $\mathcal G\colon\mathcal B\to\mathcal A$, so that $$\varphi\colon \operatorname{Hom}_{\mathcal B}(\mathcal F-,-)\overset{\sim}\Rightarrow\operatorname{Hom}_{\mathcal A}(-,\mathcal G -),$$ then can we say that a map $f\colon \mathcal FX\to Y$ is injective (surjective) if and only if $\varphi(f)$ is injective (surjective)?
For example, by Frobenius reciprocity, $\operatorname{Ind}^G_H-\cong \operatorname{Res}_H^G-$. Does this mean that a map $$f\colon \Bbb FG\otimes V\to W $$ in $\Bbb FG$-$\operatorname{Mod}$ is injective if and only if the corresponding map of $\Bbb FH$-modules $$\tilde f\colon V\to \operatorname{Res}^G_HW $$ is injective? Here, $\tilde f(v)=f(1\otimes v).$ 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 $\{x,y\} \to G(\mathbb{Z})$ defined by sending $x$ and $y$ to $1$ and $-1$ respectively. This is injective.
Its transpose is a map $F(\{x,y\}) \to \mathbb{Z}$. But $F(\{x,y\}) \cong \mathbb{Z} \oplus \mathbb{Z}$, 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 $\{x\} \to G(\mathbb{Z})$ 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.