Characterization of pullback via an exact sequence

Question

I'm trying to prove the following lemma from "A course on homological algebra" by Hilton and Stammbach:

In the category of modules $B \xleftarrow{\beta} Y \xrightarrow{\alpha} A$ is a pullback of $B \xrightarrow{\psi} X \xleftarrow{\phi}A$ if and only if the sequence $0 \to Y \xrightarrow{(\alpha,\beta)} A \oplus B \xrightarrow{\phi \oplus (-\psi)} X$ is exact.

I understand how the proof in the book shows that if $Y$ is the mentioned pullback then the sequence is exact on $A\oplus B$, but I don't see how their proof shows that the sequence is exact in $Y$.

They basically show that "the universal property of $\ker \phi \oplus (-\psi)$ is the same as the universal property of the pullback $Y$".

Any help would be appreciated.

Answer 1

Let me first introduce you to some notation which I believe makes everything easier.

The arrow $X \xrightarrow{(x,y)} A \oplus B$ will be annotated $\left(\begin{matrix}x\\ y\end{matrix}\right)$ as a column matrix, and the arrow $A\oplus B \xrightarrow{(a\oplus b)} Z$ will be annotated $\left(\begin{matrix}a & b\end{matrix}\right)$ as a row matrix. With this notation, composition behaves like matrix multiplication. So importantly $$\left(\begin{matrix}a & -b\end{matrix}\right) \left(\begin{matrix}x\\ y\end{matrix}\right) = a\circ x - b\circ y$$

In the case it's equal to $0$, we would then have $a\circ x = b\circ y$.

---

If $\ker\ \left(\begin{matrix}\varphi & -\psi\end{matrix}\right) = \left(\begin{matrix}\alpha\\ \beta\end{matrix}\right)$ then by definition of kernel

• $\left(\begin{matrix}\varphi & -\psi\end{matrix}\right)\ \left(\begin{matrix}\alpha\\ \beta\end{matrix}\right) = 0$
• $\forall x,y.\ \left(\begin{matrix}\varphi & -\psi\end{matrix}\right)\left(\begin{matrix}x\\ y\end{matrix}\right) = 0 \rightarrow \exists!z.\ \left(\begin{matrix}\alpha\\ \beta\end{matrix}\right)\circ z = \left(\begin{matrix}x\\ y\end{matrix}\right)$

which is to say respectively that

• $\varphi\circ\alpha = \psi\circ\beta$
• $\forall x,y.\ \varphi\circ x = \psi\circ y \rightarrow\exists!z.\ \alpha\circ z = x\ \land\ \beta\circ z = y$

Therefore $\ker\ \left(\begin{matrix}\varphi & -\psi\end{matrix}\right) = \left(\begin{matrix}\alpha\\ \beta\end{matrix}\right)$ if and only if $(\alpha,\beta) = \operatorname{pb}\ (\varphi,\psi)$.

---

Why is this enough?

Remember that any arrow $f$ can be factored uniquely (up to isomorphism) as $f = (\operatorname{coim} f) \circ (\operatorname{im} f)$, where $\operatorname{coim}=\operatorname{coker}\circ\ker$ and $\operatorname{im}=\ker\circ\operatorname{coker}$. In particular $\operatorname{coim}$ is epic and $\operatorname{im}$ is monic.

So by uniqueness of the factorization $f$ is monic if and only if $f=\operatorname{im} f$ (and $\operatorname{coim} f = 1$).

Furthermore, if $f=\ker\ g$ for some $g$ we must have $f$ being monic, and so equal to its image.

---

Summarizing,

$(\alpha,\beta) = \operatorname{pb}\ (\varphi,\psi)$

iff

$\ker\ \left(\begin{matrix}\varphi & -\psi\end{matrix}\right) = \left(\begin{matrix}\alpha\\ \beta\end{matrix}\right)$

iff

$\left(\begin{matrix}\alpha\\ \beta\end{matrix}\right)\ =\ \operatorname{im}\ \left(\begin{matrix}\alpha\\ \beta\end{matrix}\right)\quad \land\quad \ker\ \left(\begin{matrix}\varphi & -\psi\end{matrix}\right)\ =\ \operatorname{im}\ \left(\begin{matrix}\alpha\\ \beta\end{matrix}\right)$

iff

$0 \to Y \xrightarrow{(\alpha,\beta)} A \oplus B \xrightarrow{\phi \oplus (-\psi)} X$ is exact.