How to think about the canonical map $i^{-1}i_*\mathcal{F} \to \mathcal{F}$ and similar maps of sheaves defined by adjunction Let $i \colon Z \to X$ be the inclusion of a closed subset $Z$ into $X.$ Let $\mathcal{F}$ be a sheaf on $Z.$ I would like to show that the canonical map $i^{-1}i_*\mathcal{F} \to \mathcal{F}$ is an isomorphism. (This is part of the proof of Stacks Project, Lemma 6.32.1.) There are a few things that are not clear to me, so perhaps I will bold them as I go along.
I understand that
$$(i^{-1}i_*\mathcal{F})_z = (i_*\mathcal{F})_z = \mathcal{F}_z.$$
However, we know that just because two sheaves have isomorphic stalks does not necessarily mean that the two sheaves are isomorphic. We need to show that the above isomorphism on the stalks is induced by a morphism $i^{-1}i_*\mathcal{F} \to \mathcal{F}.$ Is there a reason why the isomorphism on stalks is induced by the canonical map $i^{-1}i_*\mathcal{F} \to \mathcal{F}?$ What even does the canonical map look like?
So if I try to unpack the definitions, I get the following. Suppose $U$ is an open set of $X,$ and so $U \cap Z$ is an open set of $Z.$ Then $i^{-1}i_*\mathcal{F}(U \cap Z)$ is the sheaf associated to
\begin{align*}
U \cap Z
&\mapsto \varinjlim_{V \supseteq U\cap Z}i_*\mathcal{F}(V)\\
&= \varinjlim_{V \supseteq U\cap Z} \mathcal{F}(i^{-1}(V))\\
&= \varinjlim_{V \supseteq U\cap Z} \mathcal{F}(V \cap Z).
\end{align*}
But the last limit is just taken over all the open sets of $Z$ that contains $U \cap Z,$ so does that mean that this is exactly $\mathcal{F}(U \cap Z)?$ Then by the universal proeprty of sheafification, does this directly show that $i^{-1}i_*\mathcal{F} = \mathcal{F}?$
Another way I tried to think about this is the following. We know that we have a bijection of sets $$\operatorname{Hom}(i^{-1}i_*\mathcal{F},\mathcal{F}) \cong \operatorname{Hom}(i_*\mathcal{F},i_*\mathcal{F}).$$ I would like to guess that our canonical map $\colon i^{-1}i_*\mathcal{F} \to \mathcal{F}$ is induced by the identity map $i_*\mathcal{F} \to i_*\mathcal{F}.$ But does this identification preserve the stalks? More generally, if we have an identification between $\phi \colon f^{-1}\mathcal{G} \to \mathcal{F}$ and $\psi \colon \mathcal{G} \to f_*\mathcal{F},$ where $\mathcal{F}$ is a sheaf on $X$ and $\mathcal{G}$ is a sheaf on $Y.$ If $f(x) = y,$ is there some sort of correspondence between $\phi_x$ and $\psi_y?$
And finally, in case the answers to my previous questions won't answer this, how should I think about the other canonical maps defined by adjunction (for example, $i^*i_*\mathcal{F} \to \mathcal{F}$ and $\mathcal{G} \to i_*i^*\mathcal{G}$ for sheaves of modules)?

Edit: I didn't want to expand on my last question because that would make this post a bit too long, but perhaps it is better to include it for the sake of completeness. In particular, let us consider the map $i_*i^*\mathcal{G} \to \mathcal{G}.$ Let $\mathcal{H} = i^*\mathcal{G}.$ Then $\mathcal{H}$ is the sheafification of the presheaf $\mathcal{H}^{\text{pre}}$ defined by
$$V \mapsto \mathcal{O}_Z(V) \otimes_{i^{-1}\mathcal{O}_X(V)} \varinjlim_{W \supseteq i(V)}\mathcal{G}(W).$$
Suppose $U$ is an open set of $X.$ Let $V = i^{-1}(V) = U \cap Z.$ Then $i_*i^*\mathcal{G}(U)$ is just $\mathcal{H}(V).$ But note that $U \supseteq i(V),$ and so we have a canonical morphism $$\mathcal{G}(U) \to \varinjlim_{W \supset i(V)}\mathcal{G}(W).$$ But then we also have canonical morphisms $$\varinjlim_{W \supset i(V)}\mathcal{G}(W) \to \mathcal{O}_Z(V) \otimes_{i^{-1}\mathcal{O}_X(V)} \varinjlim_{W \supseteq i(V)}\mathcal{G}(W) = i_*\mathcal{H}^{\text{pre}}(U) \to i_*\mathcal{H}(U).$$ (Is everything correct up to here?) So this defines a map $\mathcal{G} \to i_*i^*\mathcal{G}.$ And if we were to check the stalks, can we just follow the above composition of maps (at the level of stalks) because stalks behave nicely with direct limits and tensor products?
 A: General fact: If $F$ is left adjoint to $G$, then the counit $FG \to \mathrm{id}$ is an isomorphism iff $G$ is fully faithful.
So you just have to prove that $i_*$ is fully faithful, which is easy (just use that the open subsets of $Z$ are exactly the $U \cap Z$ for open subsets $U$ of $ X$). Notice that you don't have to think about $i^{-1}$ at all.
A: Question: "Is there a reason why the isomorphism on stalks is induced by the canonical map $i^{−1}i_∗F→F$? What even does the canonical map look like?"
Answer: If $V \subseteq Z$ is an open subscheme there is a canonical map
$$i^{-1}i_*E(V):=lim_{i(V) \subseteq U}E(i^{-1}(U)) \rightarrow E(V)$$
since $V \subseteq i^{-1}(U)$. Hence there is a canonical map of sheaves
$$\phi: i^{-1}i_*E \rightarrow E.$$
In general if $f: X \rightarrow Y$ is a map of topological spaces and $x \in X$ it follows for any $E \in sh(Y)$ there is an isomorphism $f^{-1}(E)_x \cong E_{f(x)}$.
Hence
$$ i^{-1}(i_*E)z=(i_*E)_{i(z)} \cong E_z,$$
for any $z\in Z$.
When you pass to the fiber of the map $\phi$ at $z\in Z$ you get an isomorphism
$$\phi_z: (i^{-1}i_*E)_z \cong E_z \rightarrow^{\phi_z} E_z.$$
Given an element $\{e_U\}\in i^{-1}i_*E(V)$ with $i(V) \subseteq U$ it follows
$$\phi_V(e_U):=(e_U)_V \in E(V),$$
hence for an element in the stalk $(e,W)\in i^{-1}i_*E_z$, you may choose an element $\{e_U\}\in i^{-1}i_*E(V)$ representing $(e,W)$ and you get
$$\phi_z(e,W):=((e_U)_V,V)\in E_z.$$
The you should check that this map is an isomorphism.
Hence the canonical map $\phi$ is an isomorphism at the stalks at $z$ for any $z\in Z$ and is therefore an isomorphism.
