Let $A$ be a local Noetherian ring and $M$ be an $A$-module. Let $\mathfrak{a}$ be an ideal of $A$ generated by a regular $M$-sequence $s_1,\cdots,s_r$. Let $K_\bullet(s_1,\cdots,s_r;M)$ be the Koszul complex and define the cohomology $H^i(\mathfrak{a};M):=H_{r-i}(K_\bullet(s_1,\cdots,s_r;M))$. Concretely speaking, $H^i(\mathfrak{a};M)$ is defined to be the cohomology of the differentials $$\bigwedge\limits^{r-i+1}A^r\otimes M\to\bigwedge\limits^{r-i}A^r\otimes M\to\bigwedge\limits^{r-i-1}A^r\otimes M$$, where the morphism $\bigwedge\limits^{j}A^r\otimes M\to\bigwedge\limits^{j-1}A^r\otimes M$ is defined to be $$\bigwedge\limits^j X_1\wedge\cdots\wedge X_j\otimes m\mapsto\sum\limits_{p=1}^j\bigwedge\limits^{j-1}X_1\wedge\cdots\wedge\widehat{X_p}\wedge\cdots\wedge X_j\otimes s(X_p)m$$ with $X_1,\cdots,X_j\in A^r$ and $s:A^r\to A$ is defined as $s(a_1,\cdots,a_r)=s_1a_1+\cdots+s_ra_r$.
Then one may find that $H^0(\mathfrak{a};M)=\mathrm{Hom}(A/\mathfrak{a},M)$. This extends to morphisms $\psi_M^i:\mathrm{Ext}^i(A/\mathfrak{a},M)\to H^i(\mathfrak{a},M)$ for $i\geq0$, by Cartan-Eilenberg's Homological Algebra Proposition III 5.2. Why are such $\psi_M^i$'s surjective by the regularity of the sequence $s_1,\cdots,s_r$?
