# Proof of First Isomorphism Theorem for $R$-Modules

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.

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.

• 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! Commented Aug 10, 2023 at 22:37