On proving that if a commuting rectangle's right and outer square are pullbacks, then so is the left square.

Question

I'm trying to prove that if a commutative rectangle

\begin{array} \ p \xrightarrow{i} p' \xrightarrow{m} a \\ \downarrow^{j} \quad \downarrow^k \quad \downarrow^o \\ d\ \xrightarrow{l} b \xrightarrow{n} c \end{array}

for some category $\mathcal{C}$ verifies that both the 'outer' rectangle and the right square are pullbacks, then the left square is a pullback. There other direction is also true, and I've already proved it, but this implication seems a bit trickier. So far, I've done the obvious: let $x \in \mathcal{C}$ be an object with

\begin{array} \ x \xrightarrow{\phi} p' \\ \downarrow^{\psi} \quad \downarrow^k \\ d \xrightarrow{l} b \end{array}

a cone with summit $x$, i.e. a pair $(x \xrightarrow{\phi} p',x \xrightarrow{\psi} d)$ with $k \phi = l \psi$. To conclude, I'd have to show that both $\phi$ and $\psi$ factor trough a unique morphism $x \xrightarrow{h} p$.

Since the outer square is a pullback, composing $\phi$ with $m$ gives a cone over $d \xrightarrow{nl} c \xleftarrow{o} a$, and since $p$ is a pullback, we have a unique morphism $h:x \to p$ such that $m\phi = mih$ and $\psi = jh$. Thus, it would be sufficient to see that $\phi = ih$. However, $m$ need not be a monomorphism, so I presume that to show this I should use that the right square is a pullback as well.

Any hints? Thanks in advance.

Answer 1

Okay, so, I think I've got it by using the hint provided by Nex in the comments. First, I proved the actual hint,

Proposition. If the following diagram is a pulback,

\begin{array} \ p' \xrightarrow{m} a \\ \downarrow^k \quad \downarrow^o \\ b \xrightarrow{n} c \end{array}

and $f,g : y \to p'$ are such that $mf = mg$ and $kf = kg$, then $f = g$.

Proof. By hypothesis, the cones with summit $y$ given by $(mf, kf)$ and $(mg, kg)$ are the same, and thus there exists a unique morphism $q$ such that $mf = mq = mg$ , and $kf = kq = kg$. Since both $f$ and $g$ verify this, we have that $f = q = g$.

Now having proved that, and since I had already seen that $m\phi = mih$, it will be sufficient to prove that $k\phi = kih$. By commutativity of the rectangle, we know that $ki = lj$ and so, in effect,

$$ kih = ljh \stackrel{(*)}= l\psi \stackrel{(**)}= k\phi $$

with $(*)$ proven in the original question, and $(**)$ given by the fact that $(\phi, \psi)$ are a cone and in this setting, a cone over $d \xrightarrow{l} b \xleftarrow{k} p'$ implies $l\psi = k \phi$.