The following diagram of $R$-modules is commutative.
$\require{AMScd}$
\begin{CD}
M @>f>> P\\
@VVgV @VV\varphi V\\
Q @>\psi>> N
\end{CD}
Prove: $M$ is the pullback of $\varphi$ and $\psi$ $\iff$ $0\to M\xrightarrow{(f,g)}P\oplus Q\xrightarrow{\varphi-\psi}N$ is exact.
My try:
$\boxed{\Leftarrow}$ Let $T$ be a $R$-module and $(s,t):T\mapsto P\oplus Q$ such that $(\varphi-\psi)\circ (s,t)=0$. We want to prove that there exists a unique map $h:T\to M$ such that $(f,g)\circ h=(s,t)$.
From the exactness of the sequence, we have $(\varphi-\psi)\circ (f,g)=0$. ($\star$) (Also, it seems strange that I don't use that the sequence is exact at $M$..)
$\boxed{\Rightarrow}$ I have no idea how to handle this direction.
($\star$) I now want to invoke the universal property of the kernel to prove the existence of such $h$. However, in my syllabus, this universal property is not written down. However, my syllabus states the universal property of the equalizer and shows that in the case of $R$-modules, the kernel of a $R$-module homomorphism $f:R\to S$ is an equalizer.
I would like to know if it is possible to deduce the universal property from the other one, or if there is an easy direct proof of the universal property of the kernel. I think there is one, but I have been working on this problem for too long and I can't seem to see it. (I am aware of
this question, but it has no concrete answer.)
Edit.
In my syllabus, it is proved that the equalizer of $f:X\to Y$ and the zero map is the kernel of $f$. Let $\iota:\operatorname{Ker} f\to X$ be the inclusion map. The universal property of the equalizer now gives:
For every morphism $t:T\to X$ such that $ft=0$, there exists a unique morphism $h:T\to \operatorname{Ker}f$ such that $\iota h=t$.
I want to apply this to be above exercise. We have $(\varphi-\psi)\circ (s,t)=0$ and $(\varphi-\psi)\circ (f,g)=0$, where $(s,t):T\to P\oplus Q$ and $(f,g):M\to P\oplus Q$. Applying the above gives unique morphisms $h_1:T\to \operatorname{Ker}(\varphi-\psi)$ and $h_2:M\to \operatorname{Ker}(\varphi-\psi)$ such that stuff commutes. But what I want is a morphism $T\to M$..
Best Answer
As hinted in the comments by Menezio the kernel a map $f\colon M\to N$ is the equalizer of $f$ and the trivial map $0$. This is a natural requirement as we expect from the kernel that every element $x\in\ker f$ is taken to the $0$-element in $N$ after pre-composing with the inclusion $\iota\colon\ker f\to M$. Equivalently arrow-wise: we expect $f\circ\iota=0$, i.e. that pre-composing with $\iota$ equals the trivial map.
Now, the equaliser ${\rm eq}(f,0)$ consist of an object $\ker f$ and a map $\iota\colon\ker f\to M$ such that $f\circ\iota=0\circ\iota=0$ (as composing with the trivial map is always trivial) and is universal as such. That is, for every map $k\colon K\to M$ with $f\circ k=0~\circ k=0$ there exists an unqiue map $\tilde k\colon K\to\ker f$ such that $\iota\circ\tilde k=k$.
To put it otherwise: the kernel of a map $f\colon M\to N$ is universal among maps $g\colon K\to M$ such that $f\circ g=0$, which might sound like the more usual definition. Explicitly