The Stacks project

Proof. Assume $\varphi : N \to M$ satisfies the assumptions of the lemma. Note that this implies that $\mathop{\mathrm{Ker}}(\varphi ) \subset I^ nN$ for arbitrarily large $n$. Hence by Lemma 10.51.5 we see that $\varphi$ is injection. Let $Q = M/N$ so that we have a short exact sequence

Let

be a finite free resolution of $Q$. We can choose a map $\alpha : F_0 \to M$ lifting the map $F_0 \to Q$. This induces a map $\beta : F_1 \to N$ such that $\beta \circ d_2 = 0$. The extension above is split if and only if there exists a map $\gamma : F_0 \to N$ such that $\beta = \gamma \circ d_1$. In other words, the class of $\beta$ in $\mathop{\mathrm{Ext}}\nolimits ^1_ R(Q, N)$ is the obstruction to splitting the short exact sequence above.

Suppose $n$ is a large integer such that $N/I^ nN \to M/I^ nM$ is a split injection. This implies

is still short exact. Also, the sequence

is still exact. Arguing as above we see that the map $\overline{\beta } : F_1/I^ nF_1 \to N/I^ nN$ induced by $\beta$ is equal to $\overline{\gamma _ n} \circ d_1$ for some map $\overline{\gamma _ n} : F_0/I^ nF_0 \to N/I^ nN$. Since $F_0$ is free we can lift $\overline{\gamma _ n}$ to a map $\gamma _ n : F_0 \to N$ and then we see that $\beta - \gamma _ n \circ d_1$ is a map from $F_1$ into $I^ nN$. In other words we conclude that

for this $n$.

Since we have this property for arbitrarily large $n$ by assumption we conclude that the image of $\beta$ in the cokernel of $\mathop{\mathrm{Hom}}\nolimits _ R(F_0, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(F_1, N)$ is zero by Lemma 10.51.5. Hence $\beta$ is in the image of the map $\mathop{\mathrm{Hom}}\nolimits _ R(F_0, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(F_1, N)$ as desired. $\square$