Let $f: X \to Y$ be a continuous map of topological spaces $X,Y$, let $\mathcal F$ be a sheaf on $X$ and $\mathcal G$ sheaf on $Y$. It is known that the Inverse image functor $f^{-1} : {\bf Sh}(Y) \to {\bf Sh}(X)$ and Direct image functor $f_{*} : {\bf Sh}(X) \to {\bf Sh}(Y)$ are adjoint, therefore there exist canonical morphisms of sheaves
$$\mathcal G\to f_*f^{-1}\mathcal G\quad (1),\qquad f^{-1}f_*\mathcal F\to \mathcal F\quad (2)$$
In general neither the unit map (1) nor counit (2) is isomorphism. I'm interested in the case where $f$ is injectve. What can we say about maps (1) and (2) for the cases $f$ injective and ...
(a) open inclusion $f: U \hookrightarrow Y$, i.e. $f$ continuous with open $U \subset Y$
(b) closed inclusion $f: C \hookrightarrow Y$, the same with closed $C \subset Y$
(c) injective only $f: X \hookrightarrow Y$
In most testbooks $X$ and $Y$ are additionally endowed with structure of ringed spaces and (a) respec. (b) are extended to open respec. closed immersions, but I not think that it is neccessary here to bulk up $X$ and $Y$ with this additional structure.
Question: What can we say about unit (1) & counit maps (2) in cases (a), (b) & (c)? When are they isomorphisms, when surjective or injective?
Since $\mathcal G$ and $\mathcal F$ are sheaves, it suffice to look at stalks, let $x \in X, y \in Y$. Then the the stalk of $\mathcal F$ at $x$ is
$$\mathcal F_x= \varinjlim_{U \in \mathcal{O}(X) \ : \ x \in U}\mathcal F(U) $$
About $f^{-1}\mathcal G$ two important facts are known: the stalk is easy to compute: $(f^{-1}\mathcal G)_x= \mathcal G_{f(x)}$ and if $f: U \hookrightarrow Y$ open inclusion like in (a) then $f^{-1}\mathcal G = \mathcal G \vert _U$.
For $f_*\mathcal F$ it's harder. The stalk formula says
$$ (f_*\mathcal F)_y = \varinjlim_{V \in \mathcal{O}(Y) \ : \ y \in Y}\mathcal F(f^{-1}(V)) $$
If $f: C \hookrightarrow Y$ closed inclusion as in (b), then $ (f_*\mathcal F)_y= \mathcal F_y$ if $y \in F(C)$ and zero elso. If $f: U \hookrightarrow Y$ open inclusion, then $ (f_*\mathcal F)_y= \mathcal F_y$ if $y \in f(U)$ as before, but I can't find an argument why $ (f_*\mathcal F)_y=0$ if $y$ is not in the image. It can happen that every open $V \subset Y$ intersects non trivially with $U$. Can we say else something interesting about $ (f_*\mathcal F)_y$?
Finally, let's calculate the stalks of $f_*f^{-1}\mathcal G$ and $f^{-1}f_*\mathcal F$. We obtain
$$(f_*f^{-1}\mathcal G)_y = \varinjlim_{V \in \mathcal{O}(Y) \ : \ y \in Y}f^{-1}\mathcal G(f^{-1}(V))\quad (1) \ \ \qquad (f^{-1}f_*\mathcal F)_x = \varinjlim_{V \in \mathcal{O}(Y) \ : \ f(x) \in V}\mathcal F(f^{-1}(V)) \ \ (2)$$
We conclude for (2) because $f(x)$ is obviously always contained in the image of $f$, that $(f^{-1}f_*\mathcal F)_x= \mathcal F_x$. Therefore if I'm not mistaken, then (2) should be for all cases (a), (b) & (c) an iso, yes?
For (1), if $y= f(x)$ in the image, then $(f_*f^{-1}\mathcal G)_y= \mathcal G_y$. If not, then we can at least say that for $f$ closed inclusion (b), if $y \not \in f(X)$, then $(f_*f^{-1}\mathcal G)_y=0$, so (1) is at least in this case always surjective.
Is what I wrote before true? Can we say something more about cases (a) & (c) for unit map (1)?