[proofplan] We use the refinement-stable Čech cochain model: a class for the fixed cover $\mathcal{U}$ may be represented after passing to an open refinement of $\mathcal{U}$. This is the point that makes sheaf epimorphisms usable, since sections of $\mathcal{F}''$ need only lift locally to sections of $\mathcal{F}$. After establishing exactness of the localized Čech cochain sequence, we construct the connecting homomorphism by lifting a cocycle in $\mathcal{F}''$, taking its Čech coboundary in $\mathcal{F}$, and identifying that coboundary with a cocycle in $\mathcal{F}'$. The usual elementwise diagram chase then proves exactness at every term and naturality. [/proofplan]

[step:Set up the refinement-stable Čech cochain complex] Write the open cover as $\mathcal{U} = (U_i)_{i \in I}$, where $I$ is an index set. A refinement of $\mathcal{U}$ is an open cover $\mathcal{V} = (V_a)_{a \in A}$ together with a map $\rho_{\mathcal{V}}: A \to I$ such that $V_a \subseteq U_{\rho_{\mathcal{V}}(a)}$ for every $a \in A$. For a sheaf of abelian groups $\mathcal{A}$ on $X$, an integer $q \geq 0$, and a refinement $\mathcal{V} = (V_a)_{a \in A}$, define the $q$-cochain group \begin{align*} C^q(\mathcal{V}, \mathcal{A}) := \prod_{(a_0,\ldots,a_q) \in A^{q+1}} \mathcal{A}(V_{a_0\cdots a_q}), \end{align*} where \begin{align*} V_{a_0\cdots a_q} := V_{a_0} \cap \cdots \cap V_{a_q}. \end{align*} For $c \in C^q(\mathcal{V}, \mathcal{A})$, write $c_{a_0\cdots a_q}$ for its component in $\mathcal{A}(V_{a_0\cdots a_q})$. Define the Čech coboundary map \begin{align*} \partial_{\mathcal{A},\mathcal{V}}^q: C^q(\mathcal{V}, \mathcal{A}) &\to C^{q+1}(\mathcal{V}, \mathcal{A}) \end{align*} by \begin{align*} (\partial_{\mathcal{A},\mathcal{V}}^q c)_{a_0\cdots a_{q+1}} = \sum_{r=0}^{q+1} (-1)^r c_{a_0\cdots \widehat{a_r}\cdots a_{q+1}} \big|_{V_{a_0\cdots a_{q+1}}}, \end{align*} where $\widehat{a_r}$ means that $a_r$ is omitted. The identity \begin{align*} \partial_{\mathcal{A},\mathcal{V}}^{q+1} \circ \partial_{\mathcal{A},\mathcal{V}}^q = 0 \end{align*} follows by pairing the two terms obtained by omitting $a_r$ and then $a_s$ with the terms obtained by omitting $a_s$ and then $a_r$; the signs are opposite. Let $\mathscr{C}^q(\mathcal{U},\mathcal{A})$ denote the refinement-stable $q$-cochain group, namely the direct limit of $C^q(\mathcal{V},\mathcal{A})$ over refinements $\mathcal{V}$ of $\mathcal{U}$. The coboundary maps commute with restriction to refinements, so they induce maps \begin{align*} \partial_{\mathcal{A}}^q: \mathscr{C}^q(\mathcal{U},\mathcal{A}) &\to \mathscr{C}^{q+1}(\mathcal{U},\mathcal{A}). \end{align*} Define \begin{align*} \mathscr{Z}^q(\mathcal{U},\mathcal{A}) &:= \ker(\partial_{\mathcal{A}}^q), \\ \mathscr{B}^q(\mathcal{U},\mathcal{A}) &:= \operatorname{im}(\partial_{\mathcal{A}}^{q-1}), \end{align*} with $\mathscr{B}^0(\mathcal{U},\mathcal{A}) := 0$. Then \begin{align*} \check{H}^q(\mathcal{U},\mathcal{A}) := \mathscr{Z}^q(\mathcal{U},\mathcal{A}) / \mathscr{B}^q(\mathcal{U},\mathcal{A}). \end{align*} If $u: \mathcal{A} \to \mathcal{B}$ is a morphism of sheaves of abelian groups, define the induced cochain map \begin{align*} u_*^q: \mathscr{C}^q(\mathcal{U},\mathcal{A}) &\to \mathscr{C}^q(\mathcal{U},\mathcal{B}) \end{align*} by applying $u$ componentwise on every finite intersection. Since morphisms of sheaves commute with restriction maps, one has \begin{align*} \partial_{\mathcal{B}}^q \circ u_*^q = u_*^{q+1} \circ \partial_{\mathcal{A}}^q. \end{align*} Thus $u$ induces homomorphisms on Čech cohomology. [/step]

[step:Use local lifting to obtain an exact sequence of cochain complexes]We claim that for every $q \geq 0$ the sequence \begin{align*} 0 \to \mathscr{C}^q(\mathcal{U},\mathcal{F}') \xrightarrow{f_*^q} \mathscr{C}^q(\mathcal{U},\mathcal{F}) \xrightarrow{g_*^q} \mathscr{C}^q(\mathcal{U},\mathcal{F}'') \to 0 \end{align*} is exact. Injectivity of $f_*^q$ follows because $f: \mathcal{F}' \to \mathcal{F}$ is a monomorphism of sheaves, hence $f_V: \mathcal{F}'(V) \to \mathcal{F}(V)$ is injective for every open set $V \subseteq X$. The inclusion $\operatorname{im}(f_*^q) \subseteq \ker(g_*^q)$ follows from $g \circ f = 0$. Conversely, if $b \in \mathscr{C}^q(\mathcal{U},\mathcal{F})$ satisfies $g_*^q(b) = 0$, then each component of $b$ lies in the kernel sheaf $\ker(g)$. Exactness of \begin{align*} 0 \to \mathcal{F}' \xrightarrow{f} \mathcal{F} \xrightarrow{g} \mathcal{F}'' \to 0 \end{align*} identifies $\mathcal{F}'$ with $\ker(g)$ through $f$, so there is a unique $a \in \mathscr{C}^q(\mathcal{U},\mathcal{F}')$ with $f_*^q(a)=b$. Hence $\ker(g_*^q) \subseteq \operatorname{im}(f_*^q)$. It remains to prove surjectivity of $g_*^q$. Let $c'' \in \mathscr{C}^q(\mathcal{U},\mathcal{F}'')$. Choose a representative of $c''$ on some refinement $\mathcal{V} = (V_a)_{a \in A}$. For every multi-index $(a_0,\ldots,a_q)$ and every point $x \in V_{a_0\cdots a_q}$, surjectivity of the stalk map $g_x: \mathcal{F}_x \to \mathcal{F}''_x$ gives an open neighbourhood $W \subseteq V_{a_0\cdots a_q}$ of $x$ and a section $\widetilde{c}_{a_0\cdots a_q,W} \in \mathcal{F}(W)$ such that \begin{align*} g_W(\widetilde{c}_{a_0\cdots a_q,W}) = c''_{a_0\cdots a_q}\big|_W. \end{align*} Passing to the corresponding local refinement of the finite intersections, these local sections assemble into a representative $\widetilde{c} \in \mathscr{C}^q(\mathcal{U},\mathcal{F})$ with $g_*^q(\widetilde{c})=c''$. Therefore $g_*^q$ is surjective in the refinement-stable cochain group.[/step]

[guided]The only delicate point is the last map. A sheaf epimorphism does not mean that \begin{align*} \mathcal{F}(V) \to \mathcal{F}''(V) \end{align*} is surjective for every open set $V \subseteq X$. It means that every section of $\mathcal{F}''$ lifts locally. The refinement-stable Čech complex is built precisely to remember this local nature. Let $c'' \in \mathscr{C}^q(\mathcal{U},\mathcal{F}'')$. Choose a representative on a refinement $\mathcal{V}=(V_a)_{a \in A}$, so for each ordered multi-index $\alpha=(a_0,\ldots,a_q)$ we have a section \begin{align*} c''_\alpha \in \mathcal{F}''(V_\alpha), \qquad V_\alpha := V_{a_0} \cap \cdots \cap V_{a_q}. \end{align*} Fix a point $x \in V_\alpha$. Since the sequence of sheaves is exact, the induced sequence on stalks at $x$ is exact, so the germ $(c''_\alpha)_x \in \mathcal{F}''_x$ has a preimage in $\mathcal{F}_x$. By the definition of a stalk, that preimage is represented by a section over some open neighbourhood $W \subseteq V_\alpha$ of $x$. Thus there is a section \begin{align*} \widetilde{c}_{\alpha,W} \in \mathcal{F}(W) \end{align*} such that \begin{align*} g_W(\widetilde{c}_{\alpha,W}) = c''_\alpha\big|_W. \end{align*} Doing this for all $\alpha$ and all points of $V_\alpha$ produces local lifts on an open refinement of every finite intersection. In the refinement-stable Čech group, replacing a representative by such a local refinement is part of the definition. Hence the locally chosen lifts define an element \begin{align*} \widetilde{c} \in \mathscr{C}^q(\mathcal{U},\mathcal{F}) \end{align*} with \begin{align*} g_*^q(\widetilde{c})=c''. \end{align*} This proves surjectivity of $g_*^q$ without requiring any original section map $\mathcal{F}(V_\alpha)\to\mathcal{F}''(V_\alpha)$ to be surjective.[/guided]

[step:Define the connecting homomorphism by lifting a cocycle and taking its coboundary]Fix $q \geq 0$. Define \begin{align*} \delta^q: \check{H}^q(\mathcal{U},\mathcal{F}'') &\to \check{H}^{q+1}(\mathcal{U},\mathcal{F}') \end{align*} as follows. Let $[c''] \in \check{H}^q(\mathcal{U},\mathcal{F}'')$, and choose a cocycle representative \begin{align*} c'' \in \mathscr{Z}^q(\mathcal{U},\mathcal{F}''). \end{align*} By surjectivity of $g_*^q$, choose a cochain \begin{align*} \widetilde{c} \in \mathscr{C}^q(\mathcal{U},\mathcal{F}) \end{align*} such that \begin{align*} g_*^q(\widetilde{c}) = c''. \end{align*} Since $c''$ is a cocycle and $g_*$ commutes with Čech coboundaries, \begin{align*} g_*^{q+1}(\partial_{\mathcal{F}}^q \widetilde{c}) = \partial_{\mathcal{F}''}^q(g_*^q \widetilde{c}) = \partial_{\mathcal{F}''}^q c'' = 0. \end{align*} By exactness of the cochain sequence in degree $q+1$, there is a unique cochain \begin{align*} b' \in \mathscr{C}^{q+1}(\mathcal{U},\mathcal{F}') \end{align*} such that \begin{align*} f_*^{q+1}(b') = \partial_{\mathcal{F}}^q \widetilde{c}. \end{align*} This cochain $b'$ is a cocycle, because \begin{align*} f_*^{q+2}(\partial_{\mathcal{F}'}^{q+1} b') = \partial_{\mathcal{F}}^{q+1}(f_*^{q+1} b') = \partial_{\mathcal{F}}^{q+1}\partial_{\mathcal{F}}^q \widetilde{c} = 0, \end{align*} and $f_*^{q+2}$ is injective. Therefore $b' \in \mathscr{Z}^{q+1}(\mathcal{U},\mathcal{F}')$. Set \begin{align*} \delta^q([c'']) := [b']. \end{align*}[/step]

[guided]The construction measures the obstruction to lifting a cocycle in $\mathcal{F}''$ to a cocycle in $\mathcal{F}$. Start with a class \begin{align*} [c''] \in \check{H}^q(\mathcal{U},\mathcal{F}'') \end{align*} and choose a representative \begin{align*} c'' \in \mathscr{Z}^q(\mathcal{U},\mathcal{F}''). \end{align*} The condition $c'' \in \mathscr{Z}^q$ means \begin{align*} \partial_{\mathcal{F}''}^q c'' = 0. \end{align*} By the local lifting result, after refining the cover if necessary, we can choose \begin{align*} \widetilde{c} \in \mathscr{C}^q(\mathcal{U},\mathcal{F}) \end{align*} with \begin{align*} g_*^q(\widetilde{c}) = c''. \end{align*} This lift need not be a cocycle. Its failure to be a cocycle is the cochain \begin{align*} \partial_{\mathcal{F}}^q \widetilde{c} \in \mathscr{C}^{q+1}(\mathcal{U},\mathcal{F}). \end{align*} Now apply $g_*^{q+1}$ to this failure term. Since $g_*$ commutes with the Čech coboundary, \begin{align*} g_*^{q+1}(\partial_{\mathcal{F}}^q \widetilde{c}) = \partial_{\mathcal{F}''}^q(g_*^q\widetilde{c}) = \partial_{\mathcal{F}''}^q c'' = 0. \end{align*} Thus $\partial_{\mathcal{F}}^q \widetilde{c}$ lies in the kernel of $g_*^{q+1}$. Exactness of the localized cochain sequence identifies this kernel with the image of $f_*^{q+1}$. Therefore there is a unique cochain \begin{align*} b' \in \mathscr{C}^{q+1}(\mathcal{U},\mathcal{F}') \end{align*} such that \begin{align*} f_*^{q+1}(b') = \partial_{\mathcal{F}}^q \widetilde{c}. \end{align*} Finally we check that $b'$ is a cocycle. Applying $\partial_{\mathcal{F}'}^{q+1}$ and then $f_*^{q+2}$ gives \begin{align*} f_*^{q+2}(\partial_{\mathcal{F}'}^{q+1}b') = \partial_{\mathcal{F}}^{q+1}(f_*^{q+1}b') = \partial_{\mathcal{F}}^{q+1}\partial_{\mathcal{F}}^q\widetilde{c} = 0. \end{align*} Because $f_*^{q+2}$ is injective, this forces \begin{align*} \partial_{\mathcal{F}'}^{q+1}b'=0. \end{align*} So $b'$ defines a class in $\check{H}^{q+1}(\mathcal{U},\mathcal{F}')$, and this class is $\delta^q([c''])$.[/guided]

[step:Show that the connecting homomorphism is independent of choices] First suppose the same cocycle $c''$ is lifted by two cochains $\widetilde{c}_1,\widetilde{c}_2 \in \mathscr{C}^q(\mathcal{U},\mathcal{F})$. Then \begin{align*} g_*^q(\widetilde{c}_2-\widetilde{c}_1)=0. \end{align*} Exactness in degree $q$ gives a unique cochain $a' \in \mathscr{C}^q(\mathcal{U},\mathcal{F}')$ with \begin{align*} f_*^q(a')=\widetilde{c}_2-\widetilde{c}_1. \end{align*} Let $b_1'$ and $b_2'$ be the cocycles obtained from $\widetilde{c}_1$ and $\widetilde{c}_2$. Then \begin{align*} f_*^{q+1}(b_2'-b_1') &= \partial_{\mathcal{F}}^q\widetilde{c}_2 - \partial_{\mathcal{F}}^q\widetilde{c}_1 \\ &= \partial_{\mathcal{F}}^q(\widetilde{c}_2-\widetilde{c}_1) \\ &= \partial_{\mathcal{F}}^q(f_*^q a') \\ &= f_*^{q+1}(\partial_{\mathcal{F}'}^q a'). \end{align*} Injectivity of $f_*^{q+1}$ gives \begin{align*} b_2'-b_1'=\partial_{\mathcal{F}'}^q a', \end{align*} so $b_1'$ and $b_2'$ define the same cohomology class. Next suppose $c_1''$ and $c_2''$ are cohomologous cocycles in $\mathscr{Z}^q(\mathcal{U},\mathcal{F}'')$. After passing to a common refinement, there is a cochain $s'' \in \mathscr{C}^{q-1}(\mathcal{U},\mathcal{F}'')$ for $q \geq 1$ such that \begin{align*} c_2''-c_1''=\partial_{\mathcal{F}''}^{q-1}s''. \end{align*} Choose a lift $\widetilde{s}\in\mathscr{C}^{q-1}(\mathcal{U},\mathcal{F})$ with $g_*^{q-1}(\widetilde{s})=s''$. If $\widetilde{c}_1$ lifts $c_1''$, then \begin{align*} \widetilde{c}_2 := \widetilde{c}_1+\partial_{\mathcal{F}}^{q-1}\widetilde{s} \end{align*} lifts $c_2''$, because \begin{align*} g_*^q(\widetilde{c}_2) = g_*^q(\widetilde{c}_1) + \partial_{\mathcal{F}''}^{q-1}(g_*^{q-1}\widetilde{s}) = c_1'' + \partial_{\mathcal{F}''}^{q-1}s'' = c_2''. \end{align*} Moreover \begin{align*} \partial_{\mathcal{F}}^q\widetilde{c}_2 = \partial_{\mathcal{F}}^q\widetilde{c}_1 + \partial_{\mathcal{F}}^q\partial_{\mathcal{F}}^{q-1}\widetilde{s} = \partial_{\mathcal{F}}^q\widetilde{c}_1. \end{align*} Thus the resulting connecting cocycles agree. For $q=0$, cohomologous cocycles are equal after refinement because $\mathscr{B}^0=0$, so the same argument reduces to the first case. Hence $\delta^q$ is well-defined. Additivity follows because all choices and all coboundary maps are homomorphisms of abelian groups, so $\delta^q$ is a homomorphism. [/step]

[step:Verify exactness at the three kinds of terms] Let $q \geq 0$. First consider exactness at $\check{H}^q(\mathcal{U},\mathcal{F})$. The inclusion \begin{align*} \operatorname{im}\bigl(\check{H}^q(\mathcal{U},\mathcal{F}') \to \check{H}^q(\mathcal{U},\mathcal{F})\bigr) \subseteq \ker\bigl(\check{H}^q(\mathcal{U},\mathcal{F}) \to \check{H}^q(\mathcal{U},\mathcal{F}'')\bigr) \end{align*} follows from $g\circ f=0$. Conversely, let $[c] \in \check{H}^q(\mathcal{U},\mathcal{F})$ be represented by $c \in \mathscr{Z}^q(\mathcal{U},\mathcal{F})$, and suppose $g_*[c]=0$. If $q=0$, then $g_*^0(c)=0$, so exactness of the cochain sequence gives $c=f_*^0(a')$ for some $a'\in\mathscr{C}^0(\mathcal{U},\mathcal{F}')$. Since \begin{align*} f_*^1(\partial_{\mathcal{F}'}^0a') = \partial_{\mathcal{F}}^0(f_*^0a') = \partial_{\mathcal{F}}^0c = 0, \end{align*} injectivity of $f_*^1$ gives $\partial_{\mathcal{F}'}^0a'=0$, so $[c]$ lies in the image of $\check{H}^0(\mathcal{U},\mathcal{F}')$. If $q\geq 1$, the condition $g_*[c]=0$ means that, after a common refinement, there is $s''\in\mathscr{C}^{q-1}(\mathcal{U},\mathcal{F}'')$ such that \begin{align*} g_*^q(c)=\partial_{\mathcal{F}''}^{q-1}s''. \end{align*} Choose $\widetilde{s}\in\mathscr{C}^{q-1}(\mathcal{U},\mathcal{F})$ with $g_*^{q-1}(\widetilde{s})=s''$, and define \begin{align*} d := c-\partial_{\mathcal{F}}^{q-1}\widetilde{s}. \end{align*} Then \begin{align*} g_*^q(d) = g_*^q(c)-\partial_{\mathcal{F}''}^{q-1}s'' = 0. \end{align*} By exactness of the cochain sequence, $d=f_*^q(a')$ for some $a'\in\mathscr{C}^q(\mathcal{U},\mathcal{F}')$. Since $c$ is a cocycle, \begin{align*} f_*^{q+1}(\partial_{\mathcal{F}'}^q a') = \partial_{\mathcal{F}}^q(f_*^q a') = \partial_{\mathcal{F}}^q d = \partial_{\mathcal{F}}^q c - \partial_{\mathcal{F}}^q\partial_{\mathcal{F}}^{q-1}\widetilde{s} = 0. \end{align*} Injectivity of $f_*^{q+1}$ gives $\partial_{\mathcal{F}'}^q a'=0$. Also $[c]=[d]=f_*[a']$, so exactness holds at $\check{H}^q(\mathcal{U},\mathcal{F})$. Next consider exactness at $\check{H}^q(\mathcal{U},\mathcal{F}'')$. If $[c]=g_*[a]$ with $a\in\mathscr{Z}^q(\mathcal{U},\mathcal{F})$, then in the construction of $\delta^q$ we may use the lift $\widetilde{c}=a$. Since $\partial_{\mathcal{F}}^q a=0$, the resulting cocycle in $\mathcal{F}'$ is $0$, so $\delta^q([c])=0$. Conversely, suppose $[c'']\in\check{H}^q(\mathcal{U},\mathcal{F}'')$ satisfies $\delta^q([c''])=0$. Choose a cocycle representative $c''$ and a lift $\widetilde{c}\in\mathscr{C}^q(\mathcal{U},\mathcal{F})$ with $g_*^q(\widetilde{c})=c''$. Let $b'\in\mathscr{Z}^{q+1}(\mathcal{U},\mathcal{F}')$ be defined by \begin{align*} f_*^{q+1}(b')=\partial_{\mathcal{F}}^q\widetilde{c}. \end{align*} The condition $\delta^q([c''])=0$ means that, after refinement, there is $a'\in\mathscr{C}^q(\mathcal{U},\mathcal{F}')$ such that \begin{align*} b'=\partial_{\mathcal{F}'}^q a'. \end{align*} Define \begin{align*} e := \widetilde{c}-f_*^q(a'). \end{align*} Then \begin{align*} \partial_{\mathcal{F}}^q e = \partial_{\mathcal{F}}^q\widetilde{c} - f_*^{q+1}(\partial_{\mathcal{F}'}^q a') = f_*^{q+1}(b') - f_*^{q+1}(b') = 0, \end{align*} so $e\in\mathscr{Z}^q(\mathcal{U},\mathcal{F})$. Moreover \begin{align*} g_*^q(e) = g_*^q(\widetilde{c}) - g_*^q f_*^q(a') = c''. \end{align*} Thus $[c'']$ lies in the image of $\check{H}^q(\mathcal{U},\mathcal{F})$. Finally consider exactness at $\check{H}^{q+1}(\mathcal{U},\mathcal{F}')$. For every $[c'']\in\check{H}^q(\mathcal{U},\mathcal{F}'')$, the construction of $\delta^q$ gives a lift $\widetilde{c}$ and a cocycle $b'$ satisfying \begin{align*} f_*^{q+1}(b')=\partial_{\mathcal{F}}^q\widetilde{c}. \end{align*} Hence $f_*[b']=0$ in cohomology, so \begin{align*} \operatorname{im}(\delta^q) \subseteq \ker\bigl(\check{H}^{q+1}(\mathcal{U},\mathcal{F}') \to \check{H}^{q+1}(\mathcal{U},\mathcal{F})\bigr). \end{align*} Conversely, let $[b']\in\check{H}^{q+1}(\mathcal{U},\mathcal{F}')$ be represented by $b'\in\mathscr{Z}^{q+1}(\mathcal{U},\mathcal{F}')$, and suppose $f_*[b']=0$. Then after refinement there is $a\in\mathscr{C}^q(\mathcal{U},\mathcal{F})$ such that \begin{align*} f_*^{q+1}(b')=\partial_{\mathcal{F}}^q a. \end{align*} Define \begin{align*} c'' := g_*^q(a). \end{align*} Then $c''$ is a cocycle because \begin{align*} \partial_{\mathcal{F}''}^q c'' = \partial_{\mathcal{F}''}^q(g_*^q a) = g_*^{q+1}(\partial_{\mathcal{F}}^q a) = g_*^{q+1}(f_*^{q+1}b') = 0. \end{align*} Using $a$ as the lift of $c''$ in the definition of $\delta^q$, the connecting cocycle is exactly $b'$. Therefore \begin{align*} \delta^q([c''])=[b']. \end{align*} This proves exactness at $\check{H}^{q+1}(\mathcal{U},\mathcal{F}')$. Together these three verifications give the full long exact sequence beginning with \begin{align*} 0 \to \check{H}^0(\mathcal{U},\mathcal{F}') \to \check{H}^0(\mathcal{U},\mathcal{F}) \to \check{H}^0(\mathcal{U},\mathcal{F}'') \xrightarrow{\delta^0} \check{H}^1(\mathcal{U},\mathcal{F}') \to \cdots. \end{align*} [/step]

[step:Check naturality of the construction] Let \begin{align*} (\alpha',\alpha,\alpha''): (0\to\mathcal{F}'\xrightarrow{f}\mathcal{F}\xrightarrow{g}\mathcal{F}''\to0) \to (0\to\mathcal{G}'\xrightarrow{f_{\mathcal{G}}}\mathcal{G}\xrightarrow{g_{\mathcal{G}}}\mathcal{G}''\to0) \end{align*} be a morphism of short exact sequences of sheaves of abelian groups on $X$, meaning that $\alpha':\mathcal{F}'\to\mathcal{G}'$, $\alpha:\mathcal{F}\to\mathcal{G}$, and $\alpha'':\mathcal{F}''\to\mathcal{G}''$ are sheaf morphisms satisfying \begin{align*} \alpha\circ f &= f_{\mathcal{G}}\circ \alpha', \\ \alpha''\circ g &= g_{\mathcal{G}}\circ \alpha. \end{align*} Let $[c'']\in\check{H}^q(\mathcal{U},\mathcal{F}'')$, choose a cocycle representative $c''$, and choose a lift $\widetilde{c}\in\mathscr{C}^q(\mathcal{U},\mathcal{F})$ with $g_*^q(\widetilde{c})=c''$. If $b'\in\mathscr{Z}^{q+1}(\mathcal{U},\mathcal{F}')$ is defined by \begin{align*} f_*^{q+1}(b')=\partial_{\mathcal{F}}^q\widetilde{c}, \end{align*} then $\alpha_*^q(\widetilde{c})$ is a lift of $\alpha''_*{}^q(c'')$, since \begin{align*} g_{\mathcal{G},*}^q(\alpha_*^q\widetilde{c}) = \alpha''_*{}^q(g_*^q\widetilde{c}) = \alpha''_*{}^q(c''). \end{align*} The connecting cocycle for $\alpha''_*[c'']$ is $\alpha'_*{}^{q+1}(b')$, because \begin{align*} f_{\mathcal{G},*}^{q+1}(\alpha'_*{}^{q+1}b') = \alpha_*^{q+1}(f_*^{q+1}b') = \alpha_*^{q+1}(\partial_{\mathcal{F}}^q\widetilde{c}) = \partial_{\mathcal{G}}^q(\alpha_*^q\widetilde{c}). \end{align*} Therefore \begin{align*} \delta_{\mathcal{G}}^q(\alpha''_*[c'']) = \alpha'_* \delta_{\mathcal{F}}^q([c'']). \end{align*} So the connecting maps commute with morphisms of short exact sequences. This proves naturality and completes the proof. [/step]