Given a sheaf $F:OX^{op}\rightarrow Set$ on a topological space $X$, we have the stalks $F_x:=Stk_x F:=colim_{x \in U}FU$. Then given sections $s,t \in FU$ over $U$, we have if $s_x=t_x$ then $s=t$; and then $s_y=t_y$ for all $y \in U$.

This says that sections of sheaves are locally constant when we look at their etale bundle avatar.

But sheaves with locally constant functions are called constant sheaves which are sheafications of constant presheaves (according to nlab).

This suggests that there are sheaves whose sections aren't locally constant.

Where have I gone wrong?

  • 2
    $\begingroup$ The argument in your first paragraph is missing steps. (The conclusion is wrong, at any rate.) $\endgroup$ – Zhen Lin Oct 24 '14 at 10:36

The flaw is in you claim : for $x\in U$ and $s,t \in FU$, if $s_x = t_x$ then $s=t$.

This is false. Two sections have the same image in the stalk at $x$ if and only if they coincide on some neighbourhood of $x$. That is, for $x \in U$ and $x,t\in FU$, one has $$ s_x = t_x \iff \exists W \subseteq U,\, s\!\!\restriction _W = t\!\!\restriction _W. $$

The point is that this neighbourhood depends on $s$ an $t$.

For example, take the topological space $\mathbb R$ and the sheaf $\mathcal C(-)$ of continuous real-valued functions on $\mathbb R$ (the restriction maps being the restriction of functions). Then $f \colon (-1,1) \to \mathbb R,\,y\mapsto |y|$ and $g \colon (-1,1) \to \mathbb R,\, y \mapsto y$ are two sections in $\mathcal C((-1,1))$ having the same stalk at $x=1/2$. But clearly $f \neq g$.


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