Help for the proof of Lemma for pull-backs

I am learning category theory from the book by Steve Awodey, trying to complete all the proofs, and I got stuck at one. Lemma: Given the diagram above, if the square at the right and the rectangles are pull-back diagrams, then so is the square at the left.

In the book, Awodey says that the proof is done by "diagram chase", but I encounter a difficulty.

My progress: Here is where I got so far.

Suppose there is a diagram $A \leftarrow_{\pi_1} X \rightarrow_{\pi_2} E$ such that $f\pi_1 = h'\pi_2$. To show that the left square is a pull back you need to show that there exists a unique morphism $u:X\rightarrow D$ such that $h''u = \pi_1$ and $f'u =\pi_2$.

Using the pull-back property of $D$ in the rectangle, you infer that there exists a unique morphism $x:X\rightarrow D$ such that $h''x =\pi_1$ and $g'f'x = g'\pi_2$. So $x$ looks like a good candidate.

Bearing this in mind, you use the pull-back property of $E$ in the right square to show that $\pi_2$ is the unique morphism $X \rightarrow E$ such that $h' \pi_2 = f \pi_1$.

Now, assuming that $fh''=h'f'$ , you can show \begin{equation*} h'(f'x) = fh''x = f\pi_1 \end{equation*}

and thus $f'x = \pi_2$.

This would complete the proof, but I cannot prove the highlighted assumption.

So the proof is, up to now, mostly diagram chase, but I cannot get that commutation down. Do I need to assume the square is commutative, or is there a way to infer it from the given assumptions?

• Also, if you can recommend en passant a text with a (little) more conceptual discussions but comparable depth as Awodey's, please do so! – Andrea May 7 '15 at 22:22
• I assume you mean $fh'' = h'f'$. It's common to require diagrams to be commutative, so I'd say you can just take this as one of the givens. – aes May 7 '15 at 23:10
• Thanks @aes, that's what I meant. – Andrea May 8 '15 at 9:31