Given an extension of $R$-modules $0\to B\to X\to A \to 0$, one usually associates $x\in\operatorname{Ext}^1(A,B)$ to this extension by taking the long exact sequence
$$\dotsb\to \operatorname{Hom}(A,X) \to \operatorname{Hom}(A,A) \xrightarrow{\partial} \operatorname{Ext}^1(A,B)\to \dotsb$$
and setting $x=\partial(\mathrm{id}_A)$. Alternatively one could apply $\operatorname{Ext}^{*}(-,B)$ to get
$$\dotsb\to \operatorname{Hom}(X,B) \to \operatorname{Hom}(B,B) \xrightarrow{\partial} \operatorname{Ext}^1(A,B)\to \dotsb$$
and take $y=\partial(\mathrm{id}_B)$. Do we get the same elements in this way? I.e. is $x=y$? Optimally, can you show this from the standard properties of $\operatorname{Ext}$? I became interested in this because it seems to be necessary to solve a more particular question about a proof I had.