$M/\mathfrak{m}M\cong N/\mathfrak{m}N$ implies that $M\cong N$

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Let $F$ be a field and $(A, \mathfrak{m})$ be a local $F$-algebra . Assume $f: M\to N$ be an injection of $A$-modules and $f$ induces an isomorphism $M/\mathfrak{m}M\to N/\mathfrak{m}N$, then I'd like to ask if $f$ is an isomorphism.

I have no idea of this question and I do not know how to construct an counterexample.

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It is true if $N$ is finitely generated. In fact, consider $K := \mathrm{coker}(f)$. Then $K$ is a finitely generated $A$-module and $K/\mathfrak{m}K= 0$. Hence, Nakayama's Lemma implies $K = 0$, so that $f$ is surjective.

We don't need that $A$ is an $F$-algebra.

It is not necessarily true when $N$ is not finitely generated, since there are local rings $A$ with $A$-modules $N$ such that $N/\mathfrak{m}N=0$ but $N \neq 0$, i.e. $0 \hookrightarrow N$ is a counterexample.