Example 6.29.2. Let $X = \{ s_1, s_2, \xi _1, \xi _2, \xi _3, \ldots \} $ as a set. Declare a subset $U \subset X$ to be open if $s_1 \in U$ or $s_2 \in U$ implies $U$ contains all of the $\xi _ i$. Let $U_ n = \{ \xi _ n, \xi _{n + 1}, \ldots \} $, and let $j_ n : U_ n \to X$ be the inclusion map. Set $\mathcal{F}_ n = j_{n, *}\underline{\mathbf{Z}}$. There are transition maps $\mathcal{F}_ n \to \mathcal{F}_{n + 1}$. Let $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ n$. Note that $\mathcal{F}_{n, \xi _ m} = 0$ if $m < n$ because $\{ \xi _ m\} $ is an open subset of $X$ which misses $U_ n$. Hence we see that $\mathcal{F}_{\xi _ n} = 0$ for all $n$. On the other hand the stalk $\mathcal{F}_{s_ i}$, $i = 1, 2$ is the colimit

which is not zero. We conclude that the sheaf $\mathcal{F}$ is the direct sum of the skyscraper sheaves with value $M$ at the closed points $s_1$ and $s_2$. Hence $\Gamma (X, \mathcal{F}) = M \oplus M$. On the other hand, the reader can verify that $\mathop{\mathrm{colim}}\nolimits _ n \Gamma (X, \mathcal{F}_ n) = M$. Hence some condition is necessary in part (4) of Lemma 6.29.1 above.

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