Recall that $\text{Ext}_A^1(M, N)$ is in one-to-one correspondence with equivalence classes of extensions $$0 \to N \to - \to M \to 0.$$ (Ignore Baer sums for now, use them in your answer if strictly necessary.) Moreover one can show that $\text{Ext}_A^n(M, N) = \text{Ext}(\Omega^{n-1}(M), N)$ where $\Omega^{k}(M)$ is a $k$th syzygy of $M$. Assuming as known that $\text{Ext}^1$ is independent of the given surjection from a projective, by Schnauel's lemma an isomorphism $$\text{Ext}_A^1(\Omega^{n-1}(M), N) \cong \text{Ext}_A^1(\Omega^{n-1}(M) \oplus Q, N)$$ for any projective $Q$ proves that $\text{Ext}^n$ is independent of the given projective resolution. For any short exact sequence $$0 \to N \to S \to \Omega^{n-1} \oplus Q \to 0$$ it can be shown that $S \cong K \oplus Q$ for some module $K$ and that the given sequence is equivalent to one of the form $0 \to N \to K \oplus Q \to \Omega^{n-1} \oplus Q \to 0$. How does one derive the isomorphism from this? It is claimed that the isomorphism follows directly from this, however all this tells me is that any element of $\text{Ext}_A^1(\Omega^{n-1}(M) \oplus Q, N)$ is represented by a short exact sequence with middle term having $Q$ as a direct factor.
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The last short exact sequence you write is the identity on $Q\to Q$, right? – Kevin Carlson Jan 05 '16 at 15:37
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Yes, that is correct. – Andrew Thompson Jan 05 '16 at 16:16