In the coherent world we have the following: if X is reduced and F is a coherent sheaf on it, then if the rank of all fibres in constant then F is locally free.
I thought something similar held in the topological world but cannot seem to prove it. In other words: is the following true?
Let X be a topological space (say a manifold) and consider a sheaf of Q-vector spaces F. Assume also that F(U) is finite-dimensional for all opens U. Assume also that the stalk $F_x$ has dimension n for all $x \in X$. Then F is a locally constant sheaf.
I can easily show that for each $x$ there is a neighborhood $U$, with $F(U)$ of dimension n and such that, for any $x \in V \subset U$ the restriction $F(U) \to F(V)$ is an isomorphism. However, if $V$ is a subset of U not containing x, then I don't know to show that the restriction is an isomorphism (which is what's missing to show that the sheaf is actually locally constant).
Let $X$ be a (nice) topological space, $i:Z\subset X$ a closed subset and $j:U\subset X$ the open complement. The sheaves $\mathbb{Q}_X$ and $j_!\mathbb{Q}_U\oplus i_*\mathbb{Q}_Z$ are not isomorphic in general, for instance if $Z$ is a point in $X=S^1$ (they are obviously isomorphic when $U$ and $Z$ are complementary clopens).
However, both sheaves have the same stalks, namely $\mathbb{Q}$. This comes from the fact that $j_!\mathcal{F}_x=0$ if $x\not\in U$ and $j_!\mathcal{F}_x=\mathcal{F}_x$ for $x\in U$. Similarly, $i_*\mathcal{F}_x=0$ if $x\not\in Z$ and $i_*\mathcal{F}_x=\mathcal{F}_x$ for $x\in Z$.