Exercise 1.18 of chapter II asks to show that given a map of topological spaces $f:X\to Y$, the functors $f_* : \mathsf{Sh}_Y \longrightarrow\mathsf{Sh}_X$ and $ f^{-1}: \mathsf{Sh}_X\longrightarrow \mathsf{Sh}_Y$ are adjoint. The suggestion is to define the unit and the counit and then check they satisfy the usual equations. Now I have defined the unit and counit, and want to check that if $\mathscr G$ is a sheaf on $Y$ then
$$\varepsilon_{f^{-1}\mathscr G}\circ f^{-1}(\eta_\mathscr G) : f^{-1}\mathscr G\to f^{-1}f_*f^{-1}\mathscr G\to f^{-1}\mathscr G$$
is the identity, and similarly for sheaves on $X$. This is a bit cumbersome, and the only idea I have to bypass this is perhaps to look at the maps on stalks, which I think are easily seen to be the identity because of how $f_*$, $f^{-1}$ and the unit and counit are defined. Does this suffice? If not, what is the "right" way to proceed here?
Yes: to prove two morphisms of sheaves are equal, it suffices to prove they induce the same maps on each stalk. Explicitly, let us suppose $F$ and $G$ are sheaves on a space $X$ and $a,b:F\to G$ are morphisms which are not equal. Then there is some open $U\subseteq X$ and some $s\in F(U)$ such that $a(s)\neq b(s)$. Since a section of a sheaf is determined by its germs in each stalk, $a(s)\neq b(s)$ means there exists some $x\in U$ such that $a(s)_x\neq b(s)_x$ in $G_x$. But $a(s)_x=a_x(s_x)$ and $b(s)_s=b_x(s_x)$, so this proves that $a_x\neq b_x$. That is, there is some stalk on which $a$ and $b$ are not equal.