On image of map $\text{Ext}^1_R(X,F)\to \text{Ext}^1_R(X,G)$ induced by $R$-linear map of free modules $F\to G$ with entries in the maximal ideal $\DeclareMathOperator\Ext{Ext}$Let $(R,\mathfrak m)$ be a Noetherian local ring. Let $F,G$ be finitely generated free $R$-modules and $f:F\to G$ be an $R$-linear map such that $f(F)\subseteq \mathfrak m G$. Let $X$ be a finitely generated $R$-module, and let $\alpha : 0\to F \to A_{\alpha} \to X \to 0$ be a short exact sequence i.e. $[\alpha]\in \Ext^1_R(X,F)$. We have a following push-out diagram  with $[\beta] \in \Ext^1_R(X,G)$.
$$\require{AMScd}\begin{CD}
\sigma : 0 @>>> F @>>> A_\alpha @>>> X @>>> 0 \\
@. @VVfV @VVV @| \\
\beta : 0 @>>> G @>>> A_\beta @>>> X @>>> 0.
\end{CD}$$
My question is: must it be necessarily true that $[\beta] \in \mathfrak m \Ext^1_R(X,G)$?
Some thoughts: The answer is affirmative if  $F\cong G\cong R$. Indeed, in this case, $f:R\to R$ must be given by multiplication by some $x\in R$. Since $f(F)\subseteq \mathfrak m G$, so $x\in \mathfrak m$. Then in $[\beta]=x[\alpha]\in \Ext^1_R(X,R)$, hence $[\beta]\in x\Ext^1_R(X,R)\subseteq \mathfrak m \Ext^1_R(X,G) $.
 A: This is true for any $\mathrm{Ext}$-degree (and, in fact, without many hypotheses except that $\mathfrak m$ is finitely generated and $F$ is free).
Let $(x_1,\dots,x_n)$ be generators of the maximal ideal $\mathfrak m$. Then there is a surjection $G^n \to \mathfrak m G$ given by
$$(g_1,\dots,g_n) \mapsto \sum x_i g_i.$$
Because $F$ is free, the map $F \to \mathfrak m G$ lifts to a map $F \to G^n$.
Now $\mathrm{Ext}(X,-)$ respects sums and takes the map multiplication-by-$x$ map $M \to M$ to the multiplication-by-$x$ map on $\mathrm{Ext}(X,M)$ (I'm assuming $R$ is commutative here, because otherwise $\mathrm{Ext}$ doesn't necessarily take values in $R$-modules). Therefore, the composite
$$\mathrm{Ext}(X,G)^n \cong \mathrm{Ext}(X,G^n) \to \mathrm{Ext}(X,\mathfrak m G) \to \mathrm{Ext}(X,G)$$
is given by $(w_1,\dots,w_n) \mapsto \sum x_i w_i$, taking values in $\mathfrak m \mathrm{Ext}(X,G)$.
As a result, because the map $\mathrm{Ext}(X,F) \to \mathrm{Ext}(X,G)$ factors through this map $\mathrm{Ext}(X,G^n)$, it also factors through $\mathfrak m \mathrm{Ext}(X,G)$.
