I'm trying to get my footing for projective limits by showing that the kernel of an $R$-module homomorphism can be realized as the pullback of a particular inverse system, namely:
Where $f:M \rightarrow N$ is the $R$-module hom and $g$ is the trivial map. To do this, what I believe is required is to show that I can pick maps $\alpha_1: \ker f \rightarrow 0$ and $\alpha_2:\ker f \rightarrow M$ making the square commute, then show that if there was some other $R$-module with maps into $0$ and $M$, it would factor through $\ker f$. Succinctly:
So my attempt at this is as follows, but I'd like some feedback to see if everything makes sense, and where it could be improved, because it feels very clunky. Moreover is this a statement that's supposed to be immediately obvious and I'm doing too much work? Thanks in advance for the clarification.
Attempt: The map $\alpha_1:\ker f \rightarrow 0$ is the zero map, while $\alpha_2: \ker f \rightarrow M$ is inclusion, which are both $R$-module homomorphisms. Note that $g\alpha_1(m) = 0$ for all $m \in \ker f$ and $f\alpha_2(m) = 0$ since $m \in \ker f$. Therefore the inner triangles commute. Now suppse that there is an $R$-module $D$ with maps as defined above. Note that since $g$ is the trivial map $g\beta_1 = f \beta_2$ implies that $f \beta_2(d) = 0$ for all $d \in D$ and so $\text{Im} \beta_2 \subseteq \ker f$. Define $\gamma:D \rightarrow \ker f$ by $\gamma(d) = \beta_2(d)$, it is an $R$-module homomorphism because $\beta_2$ is an $R$-module homomorphism. It is immediate that $\beta_2 = \alpha_2 \gamma$ since $\alpha_2$ includes elements of the kernel into $M$. Moreover $\beta_1$ must be the zero map and so $\beta_1 = \alpha_1 \gamma$ since $\alpha_1$ is also a zero map. Lastly $\gamma$ is unique, because if $\rho:D \rightarrow \ker f$ was another map making the entire diagram commute, then $\beta_2 = \alpha_2 \rho$ implies that $\rho(d) = \beta_2(d) = \gamma(d)$ since $\alpha_2$ is inclusion into $M$.
Best Answer
First off, I agree with HallaSurvivor!
One could, perhaps, simplify the argument by first establishing, or rather extracting, the universal property of the kernel. This way the problem transforms to checking the satisfiability of certain diagrams.
The universal property of the kernel of $f\colon M\to N$ can be stated as follows: A pair $(K,\iota\colon K\to M)$ is the kernel of $f\colon M\to N$ if $f\iota=0$ and for any $g\colon L\to M$ so that $fg=0$, there is a unique map $h\colon L\to K$ so that $g=\iota h$.
Checking that $(\ker f,\ker f\hookrightarrow M)$ satisfies this universal property can be done by hand and is essentialy contained in your argument. Indeed, working element-wise here reduces the task to a set-theoretic one. Put differently, we simply rephrase the set-theoretic definition of the kernel in arrow-theoretic terms.
Back to the pullback square.
If we let $\alpha_2\colon\ker f\hookrightarrow M$ be the canonical inclusion from the universal property and consider as $\alpha_1\colon\ker f\to0$ the zero map, the square commutes since $f\alpha_2=f\iota=0$ and any map composed with the zero map, is the zero map.
Now, given any data as necesary for the universal property of the pullback, i.e. giving maps $\beta_2\colon D\to M$ and $\beta_1\colon D\to0$ so that the square commutes, simply translates to $f\beta_2=0$, since, again, any composition with the zero map, is the zero map. The universal property of the kernel now provides a unique $\gamma\colon D\to\ker f$, such that $\alpha_2\gamma=\beta_2$. The remaining condition, that is, $\alpha_1\gamma=\beta_1$ is immediate. This is the case as any map into the $0$ module is trivial. Indeed, $\beta_1$ and $\alpha_1\gamma$ necessarily agree point-wise as there is only one possible image.
More informally, the additional side of the square does not add any information besides $f\alpha_2=f\beta_2=0$ since any composition with the zero map and any map into or out of the zero module are alway trivial, i.e. the zero map. In particular, any diagrams necessarily commute (the zero module is a zero object).