Example 17.10.9. Let $X$ be countably many copies $L_1, L_2, L_3, \ldots$ of the real line all glued together at $0$; a fundamental system of neighbourhoods of $0$ being the collection $\{ U_ n\} _{n \in \mathbf{N}}$, with $U_ n \cap L_ i = (-1/n, 1/n)$. Let $\mathcal{O}_ X$ be the sheaf of continuous real valued functions. Let $f : \mathbf{R} \to \mathbf{R}$ be a continuous function which is identically zero on $(-1, 1)$ and identically $1$ on $(-\infty , -2) \cup (2, \infty )$. Denote $f_ n$ the continuous function on $X$ which is equal to $x \mapsto f(nx)$ on each $L_ j = \mathbf{R}$. Let $1_{L_ j}$ be the characteristic function of $L_ j$. We consider the map

$\bigoplus \nolimits _{j \in \mathbf{N}} \mathcal{O}_ X \longrightarrow \bigoplus \nolimits _{j, i \in \mathbf{N}} \mathcal{O}_ X, \quad e_ j \longmapsto \sum \nolimits _{i \in \mathbf{N}} f_ j 1_{L_ i} e_{ij}$

with obvious notation. This makes sense because this sum is locally finite as $f_ j$ is zero in a neighbourhood of $0$. Over $U_ n$ the image of $e_ j$, for $j > 2n$ is not a finite linear combination $\sum g_{ij} e_{ij}$ with $g_{ij}$ continuous. Thus there is no neighbourhood of $0 \in X$ such that the displayed map is given by a “matrix” as in the proof of Lemma 17.10.8 above.

Note that $\bigoplus \nolimits _{j \in \mathbf{N}} \mathcal{O}_ X$ is the sheaf associated to the free module with basis $e_ j$ and similarly for the other direct sum. Thus we see that a morphism of sheaves associated to modules in general even locally on $X$ does not come from a morphism of modules. Similarly there should be an example of a ringed space $X$ and a quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$ such that $\mathcal{F}$ is not locally of the form $\mathcal{F}_ M$. (Please email if you find one.) Moreover, there should be examples of locally compact spaces $X$ and maps $\mathcal{F}_ M \to \mathcal{F}_ N$ which also do not locally come from maps of modules (the proof of Lemma 17.10.8 shows this cannot happen if $N$ is free).

Comment #4318 by Minh-Tien Tran on

Since $f$ is a continuous function and $f$ is defined to be zero on $(-1,1)$, it is identically zero on $[-1,1]$ as well. The same goes for $(-\infty,-2)\cup (\infty,2)$. So I believe this is the definition of $f$ you have thought of. It is not very important but can lead to unnecessary confusion.

Comment #4476 by on

The values at $-1, 1, 2, -2$ are never used. So it doesn't matter what the value is at those points. The reason for picking $(-1, 1)$ would be that it is a fixed open neighbourhood of $0$ in $\mathbf{R}$.

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