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Let $M$ be a module over a commutative and unital ring $R$. A finite presentation for $M$ is an exact sequence $$F\xrightarrow{\alpha} E\xrightarrow{\Phi} M\rightarrow 0,$$ where $E$ and $F$ are free $R$-modules with finite bases.

Suppose we are given another finite representation $$F'\xrightarrow{\alpha'} E'\xrightarrow{\Phi'} M\rightarrow 0$$ of $M$.

Claim. There are linear maps $\beta:E\rightarrow E'$, $\gamma:F\rightarrow F'$ such that $\Phi'\circ\beta=\Phi$ and $\beta\circ\alpha=\alpha'\circ\gamma.$

Hint. $\gamma$ exists because of the freeness of $F$ and the exactness at $E$ and $E'$.

My thoughts. Since $E$ and $F$ are free it is enough to definfe $\beta$ and $\gamma$ on a basis. Consider bases $(e_1,...,e_n)$ and $(f_1,...,f_m)$ of $E$ respective $F$.

Since $\Phi'$ is surjective there is, for every $i=1,...,n$, a $e'_i\in E'$ such that $\Phi'(e'_i)=\Phi(e_i)$. Thus we can define $\beta(e_i):=e'_i$ and therefore get $\Phi'\circ\beta=\Phi$.

But what about $\gamma$?

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The principle here is that if $E$ is free, $f:M\to N$ is surjective, and $h:E\to N$, then there is $h'$ with $h=f\circ h'$. This is essentially your argument at the end. As you say you apply this lemma to get $\beta$.

Now let $K$ be the kernel of $\phi'$. There's a map $\beta\circ\alpha:F\to E'$ whose image is contained inside $K$ so we can think of it as a map from $F$ to $K$ (call it $\delta$ say).

Also there is a surjection $F'\to K$ induced by $\alpha'$. Apply the same argument to lift $\delta :F\to K$ to $\beta:F\to F'$.

Iterating this argument gives a standard result in homological algebra: between any two free resolutions of a module, there is a chain map preserving the "augmentation map" to the module.

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  • $\begingroup$ Thank you! Is there a difference between $K$ and $K'$? $\endgroup$
    – user444847
    May 9, 2017 at 18:18
  • $\begingroup$ Thanks for your edit! Can you explain why $\beta\circ\alpha$ maps into $K$? $\endgroup$
    – user444847
    May 9, 2017 at 19:56
  • $\begingroup$ I got it, it is because $\text{im}\alpha=\text{ker}\Phi$ and $\Phi=\Phi'\circ\beta.$ $\endgroup$
    – user444847
    May 9, 2017 at 20:01

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