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Let $(R,\mathfrak m,k)$ be a local ring of depth $d$ and $u:F_1\rightarrow F_0$ a homomorphism of finite free modules such that $\operatorname{Im}u\subset \mathfrak mF_0$. Then this map induces the zero map $$\mathrm{Ext}^d(k,F_1)\rightarrow\mathrm{Ext}^d(k,F_0).$$ Could you explain me why?

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Take a regular sequence of length $d$ in the maximal ideal $\mathfrak m$. Then $\mathrm{Ext}^d_k(k,F_0)\cong\mathrm{Hom}_{\bar{R}}(k,\bar{F}_0)$, where the bar denotes the quotient module by the regular sequence. But $$\mathrm{Hom}_{\bar{R}}(k,\bar{F}_0)=(0:\mathfrak m)_{\bar{F}_0},$$ so the map $u$ induces a map of socles $(0:\mathfrak m)_{\bar{F}_1}\rightarrow (0:\mathfrak m)_{\bar{F}_0}$, but since the map $u$ is minimal, the matrix associated to $u$ has all entries in $\mathfrak m$, and so the induced map on the socles is zero.

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