Let \(\varphi: \mcf \to \mcg\) be a morphism of sheaves.
Show that \(\im \varphi \cong \mcf / \ker \varphi\).
Show that \(\coker \varphi \cong \mcg / \im \varphi\).
Solution
1. The morphism \(\varphi:\mcf\to\mcg\) factors canonically as\[ \mcf\xrightarrow{q}\mcf/\ker\varphi \xrightarrow{\bar\varphi}\im\varphi \hookrightarrow\mcg. \]
The kernel subsheaf \(\ker\varphi\subseteq\mcf\) is annihilated by \(\varphi\). By the universal property of the quotient sheaf, \(\varphi\) therefore induces a unique morphism \[ \bar\varphi:\mcf/\ker\varphi\longrightarrow\mcg. \] Its image is exactly the image sheaf \(\im\varphi\), so it factors uniquely through the inclusion \[ \im\varphi\hookrightarrow\mcg. \] This gives the displayed factorization.
2. The induced morphism\[ \boxed{\mcf/\ker\varphi\xrightarrow{\sim}\im\varphi} \]is an isomorphism.
At a point \(x\in X\), the stalk of the quotient is \[ (\mcf/\ker\varphi)_x \cong \mcf_x/\ker(\varphi_x) \] by Exercise II.1.6. The stalk of the image sheaf is the ordinary image of the stalk map, \[ (\im\varphi)_x=\im(\varphi_x). \] The stalk map induced by \(\bar\varphi\) is therefore the ordinary first-isomorphism-theorem map \[ \mcf_x/\ker(\varphi_x) \xrightarrow{\sim} \im(\varphi_x). \] Hence \(\bar\varphi\) is an isomorphism on every stalk and therefore an isomorphism of sheaves. This proves part (a).
3. By definition, the cokernel sheaf of \(\varphi\) is the cokernel of the inclusion\[ \im\varphi\hookrightarrow\mcg. \]
The cokernel of a morphism of sheaves is obtained by sheafifying the presheaf cokernel. Since the morphism \[ \im\varphi\hookrightarrow\mcg \] is injective, its cokernel is precisely the quotient sheaf of \(\mcg\) by the subsheaf \(\im\varphi\). The original map \(\varphi\) and the inclusion of its image have the same cokernel, because the map \(\mcf\to\mcg\) factors through \(\im\varphi\) and has image exactly that subsheaf.
4. There is a canonical isomorphism\[ \boxed{\coker\varphi\cong\mcg/\im\varphi}. \]
By step 3, the cokernel fits into the short exact sequence \[ \im\varphi\longrightarrow\mcg\longrightarrow\coker\varphi\longrightarrow0. \] Apply Exercise II.1.6 to the subsheaf \(\im\varphi\subseteq\mcg\). It identifies the terminal sheaf canonically with the quotient \[ \mcg/\im\varphi. \] This proves part (b).
5. Q.E.D.
Steps <1>1–<1>2 prove part (a), and steps <1>3–<1>4 prove part (b).