= Solution
The kernel of the restriction map
$$
\Gamma(X,\mathcal F)\longrightarrow\Gamma(X\setminus Z,\mathcal F)
$$
consists exactly of sections whose germs vanish outside $Z$, namely $\Gamma_Z(X,\mathcal F)$. This proves exactness. If $\mathcal F$ is a <flasque sheaf>, the restriction map is surjective by definition.
Now let $0\to\mathcal F_1\to\mathcal F_2\to\mathcal F_3\to0$ be exact. The <global section functor> is left exact, so a section of $\Gamma_Z(X,\mathcal F_2)$ mapping to zero lifts uniquely to a global section of $\mathcal F_1$. Its germs outside $Z$ vanish because $\mathcal F_1\to\mathcal F_2$ is injective at each <stalk of a sheaf>. It therefore lies in $\Gamma_Z(X,\mathcal F_1)$, proving exactness of the supported-section sequence.
Suppose in addition that $\mathcal F_1$ is flasque and take $s_3\in\Gamma_Z(X,\mathcal F_3)$. Surjectivity on global sections gives a lift $s_2\in\Gamma(X,\mathcal F_2)$. On $X\setminus Z$, its restriction comes from some $t_1\in\Gamma(X\setminus Z,\mathcal F_1)$ by left exactness. Extend $t_1$ to $\widetilde t_1\in\Gamma(X,\mathcal F_1)$ using flasqueness. Then $s_2-\widetilde t_1$ maps to $s_3$ and vanishes outside $Z$, proving surjectivity on the right.
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