Why is $ \mathfrak{Mod}(A_{Y}/f) $ a thick subcategory of $\mathfrak{Mod}(A_{Y})$?

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Let $f: Y \to X$ be a continuous map and $\mathfrak{Mod}(A_{Y}/f)$ be the full subcategory of $\mathfrak{Mod}(A_{Y})$ (categories of $A_{Y}$-modules, with $A$ a fixed ring) whose sheaves $\mathcal{F}$ satisfy $\mathcal{F}|_{f^{-1}(x)}$ is locally constant, for all $x \in X$.

I'm trying to prove the following statement:

  • $ \mathfrak{Mod}(A_{Y}/f) $ is a thick subcategory of $\mathfrak{Mod}(A_{Y})$

So, as defined in "Sheaves on Manifolds" (which is where this statement is from), it must be checked that for any $Y,Y',Z,Z' \in \mathfrak{Mod}(A_{Y}/f) $ and $X \in \mathfrak{Mod}(A_{Y})$, if there is an exact sequence $$ Y \to Y' \to X \to Z' \to Z $$ then $X \in \mathfrak{Mod}(A_{Y}/f)$

But, since the sequence is exact, so is $ Y\vert_{f^{-1}(x)} \to Y'\vert_{f^{-1}(x)} \to X\vert_{f^{-1}(x)} \to Z'\vert_{f^{-1}(x)} \to Z\vert_{f^{-1}(x)}$

Since the restrictions are locally constant sheaves, (let us assume $f^{-1}(x)$ is connected) they have the same stalk at each point. This implies that the locally constant sheaf $M_{f^{-1}(x)}$ (where $M := X_{y}$, for some chosen $y \in f^{-1}(x)$), when sustituted for $X\vert_{f^{-1}(x)}$ preserves the exactness of the sequence. If we had a map $M_{f^{-1}(x)} \to X\vert_{f^{-1}(x)}$, we could use the five lemma do conclude the proof. But in general, I don't see a way to construct this map. And from what I know, the locally constant sheaf isn't fully determined by its stalks (right?). That is, at least in general topological spaces.

  • Am I going wrong somewhere in my reasoning? How is that subcategory thick in relation to $\mathfrak{Mod}(A_{Y})$?

Edit: in truth, we could drop the restriction to the fibers of $f$, since the question seems to have the same complexity as that of:

  • Is the category of constant sheaves $\mathfrak{Const}(A_{Y})$ a thick subcategory of $\mathfrak{Mod}(A_{Y})$?

Thank you for all the help in advance :).

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So, let us assume $Y$ is locally connected and suppose we have an exact sequence

$$ 0 \to N_{Y} \to \mathcal{F} \to C_{Y} \to 0$$ where $N,C$ are $A$-modules.

We have a morphism of chain complexes:

$\require{AMScd}$ \begin{CD} 0 @>>> N @>>> \mathcal{F}(U) @>>> C @>>> 0 \tag1\\ @V V V @VV id V @VV \rho_{U,V} V @VV id V @VV V\\ 0 @>>> N @>>> \mathcal{F}(V) @>>> C @>>> 0 \end{CD}

where we choose $U$ and $V$ to be both connected.

Let $c \in C$. By the exactness of the above sequence, we can choose for each $x \in U$, a neighbourhood $x \in U_{x} \subset U$ and a section $s^{(x)}$ of $\mathcal{F}$ over $U_{x}$ such that $s^{(x)}_{x} \mapsto c $. By the commutative diagram

$\require{AMScd}$ \begin{CD} \mathcal{F}(U) @>>> C\\ @V V V @VV id V\\ \mathcal{F}_{x} @>>> C \end{CD}

we conclude that $s^{(x)} \mapsto c$. Likewise, $s^{(y)} \mapsto c$ and so $s^{(y)}\vert_{U_{x} \cap U_{y}} = s^{(x)}\vert_{U_{x} \cap U_{y}}$ as can be checked in each connected component of the intersection, using the same reasoning as before. Then, we can glue these section to form a section $s \in \mathcal{F}(U)$ satisfying $s \mapsto c$, from which we conclude that both of the sequences in $(1)$ are exact and by the five lemma, $\rho_{U,V}$ will be an isomorphism. But then, for all connected open sets $W$, we will have $\mathcal{F}(W) \simeq M$ for some $A$-module $M$. From this we can conclude $\mathcal{F} \simeq M_{Y}$ if $Y$ is connected. If $Y$ is not connected, we can conclude such isomorphisms for each connected component, assuring at least that $\mathcal{F}$ is locally constant.

What do you think of this reasoning? Did I make any mistake somewhere?