Is the Quotient Sheaf Isomorphic to the Image Sheaf?

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Here is this week's POTW:

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Let $\mathscr{F} \overset{\eta}{\to} \mathscr{G}$ be a morphism of sheaves over a topological space $X$. Prove that quotient sheaf $\mathscr{F}/\operatorname{ker}(\eta)$ is isomorphic to the image sheaf $\operatorname{im}(\eta)$.-----

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No one answered this week's problem. You can read my solution below.

Spoiler

It suffices to prove that for every $x\in X$, the stalks $\mathscr{F}/\operatorname{ker}(\eta)$ and $\operatorname{im}(\eta)$ at $x$ are isomorphic. Fix $x\in X$. The morphism $\eta$ induces a morphism $\eta_x : \mathscr{F}_x \to \mathscr{G}_x$ on stalks. There is an isomorphism $F_x/\operatorname{ker}(\eta_x) \cong \operatorname{im}(\eta_x)$. On the other hand, $\mathscr{F}_x/\operatorname{ker}(\eta_x) \cong \left(\mathscr{F}/\operatorname{ker}(\eta)\right)_x$ and $\left(\operatorname{im}(\eta)\right)_x = \operatorname{im}(\eta_x)$. So $\left(\mathscr{F}/\operatorname{ker}(\eta)\right)_x \cong \left(\operatorname{im}(\eta)\right)_x$. Since $x$ was arbitrary, the result follows.