Let $(X,\mathcal{O}_X)$ be a ringed space.
Let $\mathcal{S}$ be a sheaf of $\mathcal{O}_X$-modules on $X$.
$\mathcal{S}$ is free if it is isomophic to a direct sum of copies
of $\mathcal{O}_X$.$\mathcal{S}$ is locally free if $X$ can be covered by open sets
$U$ for which $\mathcal{S}|_U$ is a free $\mathcal{O}_X|_U$ module.
Is there important example?
Or toy example?
Or loccaly free but not free case ?
Best Answer
Take $X$ a topological manifold and $\mathcal O_X$ its sheaf of continuous real-valued functions. Take a finite degree covering map $p : Y \to X$ and let $\mathcal S(U)$ consist of the continuous real-valued functions on $p^{-1}(U)$. (I.e. $\mathcal S = p_* \mathcal O_Y$.)
Then $\mathcal S$ is locally free, and it is free iff the covering map is trivial, i.e. when $Y$ is the product of $X$ with a discrete space.
Proof of free $\implies$ trivial cover: There are many ways to write this down; here's one which I think is most helpful:
Let $n = \dim_{\mathcal O_X(X)} \mathcal O_Y(Y)$. Take a basis $(f_i)$ of $\mathcal O_Y(Y)$ consisting of orthogonal idempotents, and let $Y_{f_i} = \{y \in Y : f_i(y) \neq 0 \} = \{ y \in Y : f_i(y) = 1 \}$. Because the $\mathcal O_Y(Y)$-ideal generated by the $f_i$ is $(1)$, $$\bigcup_i Y_{f_i} = Y_1 = Y$$ When $i \neq j$, because $f_i f_j = 0$, $$Y_{f_i} \cap Y_{f_j} = Y_0 = \varnothing$$ So that the $Y_{f_i}$ form a disjoint cover of $Y$. For each $i$ we have the restriction morphism $\mathcal O_Y(Y) \to \mathcal O_Y(Y_{f_i})$ which induces an isomorphism $\mathcal O_Y(Y)_{f_i} \cong O_Y(Y_{f_i})$.
Call $p_i$ the restriction of $p$ to $Y_{f_i}$. We show that each $p_i$ is a topological isomorphism. By assumption, we have an isomorphism $$\begin{align*} \mathcal O_X(X)^n &\to \mathcal O_{Y}(Y) \\ (g_i) &\mapsto \sum_{i=1}^n g_i \circ p \cdot f_i \end{align*}$$ By localizing, for each $i$ this induces an isomorphism $$\begin{align*} \mathcal O_X(X) & \to O_{Y}(Y)_{f_i} \to \mathcal O_{Y}(Y_{f_i}) \\ g &\mapsto g \circ p \cdot f_i \mapsto g \circ p_i \cdot f_i|_{Y_{f_i}} = g \circ p_i \end{align*}$$
We show that each $p_i$ is surjective by using that the above map is injective: Assume $X$ and $Y$ are compact, then the argument is nicer: The set $Y_{f_i}$ is closed, hence compact. Suppose the closed set $p_i(Y_{f_i})$ does not equal $X$. Let $0 \neq g \in \mathcal O_X(X)$ be such that $X_g \subset X - p_i(Y_{f_i})$. Then $g \circ p_i = 0$, contradicting the injectivity. Thus $p_i$ is surjective.
In the general case, injectivity of $g \mapsto g \circ p_i$ shows that each $p_i(Y_{f_i})$ is dense in $X$. Let $x \in X$. By assumption, there exists $m \in \mathbb N$ and $U \subset X$ open, containing $x$, such that $p^{-1}(U)$ is a disjoint union of $m$ open sets $V_j$, such that the restriction of $p$ to each is a homeomorphism. Take $U$ connected. Then the $V_j$ are connected. Because the connected components of $Y$ are the $Y_{f_k}$, for every $j$ there exists $k(j)$ with $V_j \subset Y_{f_{k(j)}}$. Now because $p_i(Y_{f_i})$ is dense in $X$, it intersects $U$, so $Y_{f_i}$ must intersect one one the $V_j$. Then $i = k(j)$, so $$x \in U = p_i(V_j) \subset p_i(Y_{f_i})$$ so that $p_i$ is surjective. (As a corollary, $m = n$ does not depend on $U$, so that $\deg p = \dim_{\mathcal O_{X}} \mathcal O_Y$.)
Finally, that $p_i$ is a homeomorphism follows from the fact that it is a degree 1 covering map. Or alternatively, because $\mathcal O_X(X) \cong \mathcal O_{Y}(Y_{f_i})$, and the topologies on $X$ and $Y_{f_i}$ are induced by the open sets $X_g$ resp. $(Y_{f_i})_f$ for $g \in \mathcal O_X(X)$ resp. $f \in \mathcal O_Y(Y_{f_i})$. (Use bump functions.)