We first show that if $\mathcal{K}$ is an injective $\mathcal{O}_X$-module, then there does exist a sequence (\ref{equation-extension}) with $c_{\mathcal{F}'} = c$. To do this, choose a flat $\mathcal{O}_{X'}$-module $\mathcal{H}'$ and a surjection $\mathcal{H}' \to \mathcal{F}$ (Modules, Lemma \ref{modules-lemma-module-quotient-flat}). Let $\mathcal{J} \subset \mathcal{H}'$ be the kernel. Since $\mathcal{H}'$ is flat we have $$ \mathcal{I} \otimes_{\mathcal{O}_{X'}} \mathcal{H}' = \mathcal{I}\mathcal{H}' \subset \mathcal{J} \subset \mathcal{H}' $$ Observe that the map $$ \mathcal{I}\mathcal{H}' = \mathcal{I} \otimes_{\mathcal{O}_{X'}} \mathcal{H}' \longrightarrow \mathcal{I} \otimes_{\mathcal{O}_{X'}} \mathcal{F} = \mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{F} $$ annihilates $\mathcal{I}\mathcal{J}$. Namely, if $f$ is a local section of $\mathcal{I}$ and $s$ is a local section of $\mathcal{H}$, then $fs$ is mapped to $f \otimes \overline{s}$ where $\overline{s}$ is the image of $s$ in $\mathcal{F}$. Thus we obtain $$ \xymatrix{ \mathcal{I}\mathcal{H}'/\mathcal{I}\mathcal{J} \ar@{^{(}->}[r] \ar[d] & \mathcal{J}/\mathcal{I}\mathcal{J} \ar@{..>}[d]_\gamma \\ \mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{F} \ar[r]^-c & \mathcal{K} } $$ a diagram of $\mathcal{O}_X$-modules. If $\mathcal{K}$ is injective as an $\mathcal{O}_X$-module, then we obtain the dotted arrow. Denote $\gamma' : \mathcal{J} \to \mathcal{K}$ the composition of $\gamma$ with $\mathcal{J} \to \mathcal{J}/\mathcal{I}\mathcal{J}$. A local calculation shows the pushout $$ \xymatrix{ 0 \ar[r] & \mathcal{J} \ar[r] \ar[d]_{\gamma'} & \mathcal{H}' \ar[r] \ar[d] & \mathcal{F} \ar[r] \ar@{=}[d] & 0 \\ 0 \ar[r] & \mathcal{K} \ar[r] & \mathcal{F}' \ar[r] & \mathcal{F} \ar[r] & 0 } $$ is a solution to the problem posed by the lemma. \medskip\noindent General case. Choose an embedding $\mathcal{K} \subset \mathcal{K}'$ with $\mathcal{K}'$ an injective $\mathcal{O}_X$-module. Let $\mathcal{Q}$ be the quotient, so that we have an exact sequence $$ 0 \to \mathcal{K} \to \mathcal{K}' \to \mathcal{Q} \to 0 $$ Denote $c' : \mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{F} \to \mathcal{K}'$ be the composition. By the paragraph above there exists a sequence $$ 0 \to \mathcal{K}' \to \mathcal{E}' \to \mathcal{F} \to 0 $$ as in (\ref{equation-extension}) with $c_{\mathcal{E}'} = c'$. Note that $c'$ composed with the map $\mathcal{K}' \to \mathcal{Q}$ is zero, hence the pushout of $\mathcal{E}'$ by $\mathcal{K}' \to \mathcal{Q}$ is an extension $$ 0 \to \mathcal{Q} \to \mathcal{D}' \to \mathcal{F} \to 0 $$ as in (\ref{equation-extension}) with $c_{\mathcal{D}'} = 0$. This means exactly that $\mathcal{D}'$ is annihilated by $\mathcal{I}$, in other words, the $\mathcal{D}'$ is an extension of $\mathcal{O}_X$-modules, i.e., defines an element $$ o(\mathcal{F}, \mathcal{K}, c) \in \Ext^1_{\mathcal{O}_X}(\mathcal{F}, \mathcal{Q}) = \Ext^2_{\mathcal{O}_X}(\mathcal{F}, \mathcal{K}) $$ (the equality holds by the long exact cohomology sequence associated to the exact sequence above and the vanishing of higher ext groups into the injective module $\mathcal{K}'$). If $o(\mathcal{F}, \mathcal{K}, c) = 0$, then we can choose a splitting $s : \mathcal{F} \to \mathcal{D}'$ and we can set $$ \mathcal{F}' = \Ker(\mathcal{E}' \to \mathcal{D}'/s(\mathcal{F})) $$ so that we obtain the following diagram $$ \xymatrix{ 0 \ar[r] & \mathcal{K} \ar[r] \ar[d] & \mathcal{F}' \ar[r] \ar[d] & \mathcal{F} \ar[r] \ar@{=}[d] & 0 \\ 0 \ar[r] & \mathcal{K}' \ar[r] & \mathcal{E}' \ar[r] & \mathcal{F} \ar[r] & 0 } $$ with exact rows which shows that $c_{\mathcal{F}'} = c$. Conversely, if $\mathcal{F}'$ exists, then the pushout of $\mathcal{F}'$ by the map $\mathcal{K} \to \mathcal{K}'$ is isomorphic to $\mathcal{E}'$ by Lemma \ref{lemma-inf-ext} and the vanishing of higher ext groups into the injective module $\mathcal{K}'$. This gives a diagram as above, which implies that $\mathcal{D}'$ is split as an extension, i.e., the class $o(\mathcal{F}, \mathcal{K}, c)$ is zero.