Vtyjkyuk

To be brief, here's my question: Suppose we have a ring $R$ (say commutative with unit) with an $R$-linear map $MxrightarrowfN$ where $M,N$ are finitely generated projective (FGP) $R$-modules. Suppose that the categorical kernel of $f$ in the category of FGP $R$-modules exists: denote it by $P$. Letting $kerf=min M: f(m)=0$ be the usual module-theoretic kernel, why must we have that $Pcong kerf$ (or is this not true)?

For a bit more background: I am ultimately trying to find examples of (non-obvious) categories which fail to have kernels. After some reading, it looks like FGP modules over appropriate rings form a common example. From what I've read, the usual way of showing this is to produce a morphism for which the module-theoretic kernel fails to be projective or finitely generated (e.g. see mdp's great answer here).

There's one subtle piece of this method that I don't understand, which is why the categorical kernel of such a morphism must be (isomorphic to) the module-theoretic kernel.

Here's what I know:

• For any non-trivial commutative ring $R$, the category of $R$-modules, $Rtextsf-Mod$, certainly has kernels, and the category of FGP $R$-modules is a full subcategory of $Rtextsf-Mod$


• For any category $mathscrC$ and full subcategory $mathscrD$ of $mathscrC$, a morphism $XxrightarrowfY$ of $mathscrD$ can have a kernel in $mathscrD$ but no kernel in $mathscrC$. However, if $f$ has a kernel in $mathscrC$ and that kernel is a morphism in $mathscrD$, it is also the kernel in $mathscrD$.


• If $mathscrC,D$ are as before, there are examples where a morphism $XxrightarrowfY$ of $mathscrD$ has a kernel in both $mathscrC$ and $mathscrD$, but these kernels are not isomorphic.


• If you denote the module theoretic kernel by $textker_R(f)$, and the categorical kernel of $f$ by $textker_FGP(f)$, there is a monomorphism $textker_FGP(f)hookrightarrowtextker_R(f)$ which commutes with the kernel morphisms, so you may replace $textker_FGP(f)$ by a submodule of $textker_R(f)$ if necessary.

With all this said, I think the proof (if it's true) has much more to do with the algebra at play, but I don't see how.

Any and every hint/bit of help is greatly appreciated, thank you!