I want to show this is true in the category of left $R$-modules.

Let $f:A\to B$ be a monic and let $g:B\to \text{coker}(f)$ be the cokernel of $f$. I want to show that $f$ is the kernel of $g$.

Let the inclusion $i:\text{ker}(g) \to B$ be the kernel of $g$.

Since $\text{Im}(f) = \text{Ker}(g)$, so $i: \text{Im(f)}\to B$.

Given $f$ is monic, we have an isomorphism of $A$ with $\text{Im(f)}$, and there exists $f^{-1}:\text{Im(f)} \to A$.

Now $gi=0$ and given any $j : C \to A$ such that $gj=0$, there exists a unique $g' : C \to \text{Im(f)}$ by the universal property of a kernel such that $ig'=j$, or $g'=j$, since $i$ is an inclusion map.

Now, putting it all together, consider $f^{-1}g':C \to A$. We have $f(f^{-1}g')=j$, and $f^{-1}g'$ is unique so it follows that $f$ is the kernel of $g$.

Is this correct?


This is essentially correct, but this is not very well organized and written.
First, there is a typo, $j$ goes from $C$ to $B$ (not $A$).
Second, at some point you need to say that $g\circ f = 0$, this is necessary in order to be able to conclude that $f = ker(g)$.
Third, it is false that $g' = j$ since $f$ may not be surjective. Actually you don't need it. One simply has $f(f^{-1}g') = (ff^{-1})g' = ig' = j$ with your notations.
Also, you might want to justify the uniqueness of $f^{-1}g'$. This is an easy consequence of the uniqueness of $g'$.
Last, $(f^{|})^{-1}$ would be a better notation for what you call $f^{-1}$ since $f$ may not be an isomorphism.


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