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As setup we have $R$ a ring, $f:M\rightarrow N$ a $R$-module homomorphism. Then $\exists$ a unique $R$-module homomorphism $\overline{f}:M/\text{ker }f\rightarrow \text{im }f$, which is also an isomorphism. I have the follow diagram.

$$\require{AMScd} \begin{CD} \text{ker }f @>i>> M @>f>> N @.\\ @. @VV\pi V @AAjA \\ @. M/\text{ker }f @>\overline{f}>> \text{im}f \end{CD}$$

$i:\text{ker }f\rightarrow M, m\mapsto m$

$\pi :M\rightarrow M/\text{ker }f, m\mapsto m+\text{ker }f$

$j:\text{im }f\rightarrow N, n\mapsto n$

My confusion is regarding the proof: "The universal property of the cokernel, applied to the composite $f\circ i =0$, yields a unique 𝑅‐module homomorphism". Should we not use the universal property of the cokernel on $i$, where im $i=$ker $f$? Therefore we would get a unique homomorphism from $M/$ker $f$ to $N$ as $N$ is a $R$-module.

Thanks in advance!

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1 Answer 1

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Here's a proof without using the explicit constructions of $i$, $\pi$, $j$ and whatnot, but only the universal properties of kernel and cokernel.

A cokernel $\pi$ of $i$ satisfies that $\forall g.\ g\circ i = 0 \Rightarrow \exists!h.\ h\circ\pi = g$. In our case $f\circ i=0$, therefore $\exists! h.\ h\circ\pi = f$.

I understand this is what he means with "applied to the composite $f\circ i$", although I would've personally said "applied to $f$".


For now we have $h:M/ker\ f \rightarrow N$, so we are not there yet with $\bar f$.

For this consider that $j$ is actually the kernel of the cokernel of $f$.

This is to say that $(coker\ f)\circ j = 0$ and $\forall g.\ (coker\ f)\circ g = 0\Rightarrow \exists! k.\ j\circ k = g$.

In our case $(coker\ f)\circ h\circ\pi = (coker\ f)\circ f = 0 = 0\circ\pi$. But since $\pi$ is epic you get $(coker\ f)\circ h=0$.

Then $\exists! k:M/ker\ f\rightarrow im\ f.\ j\circ k = h$. This $k$ is the $\bar f$ you want.

To prove that $\bar f$ is an isomorphism though is a tiny bit more involved.

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  • $\begingroup$ I don't fully understand your question, I feel it's a bit vague, so feel free to make a comment if this was not what you wanted. Cheers! $\endgroup$
    – Julián
    Commented Aug 10, 2023 at 22:37

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