Pullbacks - question 5.7.2 from Awodey's *Category Theory*

I have a problem with question 5.7.2 (b) from Awodey's Category Theory, it says:

Let C be a category with pullbacks.

(a) Show that an arrow $m : M → X$ in C is monic if and only if the diagram below is a pullback.

$\qquad\qquad\qquad\qquad\qquad\qquad$

Thus, as an object in C/X, m is monic iff $m × m \cong m$.

(b) Show that the pullback along an arrow $f : Y \to X$ of a pullback square over $X$,

$\qquad\qquad\qquad\qquad\qquad\qquad$

is again a pullback over $Y$. (Hint: draw a cube and use the two-pullbacks lemma.) Conclude that the pullback functor $f^*$ preserves products.

(c) Conclude from the foregoing that in a pullback square

$\qquad\qquad\qquad\qquad\qquad\qquad$

if $m$ is monic, then so is $m'$.

However I don't get what the question (b) is asking us - and to be honest, it is kind of confusing.

Well, he seems to suggest that we have to assume "the pullback along an arrow $f : Y → X$ of a pullback square over $X$" - but what is this supposed to mean exactly? He doesn't seem to "define" this in any part of the book.

Also, I am able to solve (c) directly. But how can we "conclude [it] from the foregoing"? How are those problems related?

Thanks!

• You have to construct a cube, given the face and the addition edge $Y\to X$. So first pullback $A\to X$ along $Y\to X$ and you'll have two faces of the cube, and actually 10 of edges, then pullback $B\to X$, etc. Once you've gotten all the edges of the cube, show that the opposite face is a pullback square. – ziggurism Nov 27 '15 at 3:18
• For sure you want to add to the square in the second question (with corners being $X, A, B$ and $A\times_X B$) an arrow $f:Y\to X$. What's eventually asked for is supposed to be a pullback square over $Y$. Ask yourself: What are ways in which the 3 remaining corners could be obtained from the 5 you already got? For the last part, think about what you start with (the diagram in that question) and what you're supposed to use (the diagram in the first question). Use the result of the second to bridge the two. – Nikolaj-K Nov 27 '15 at 3:21
• For part c, put $m$ also as the "out of the page" arrow to make your cube. Then the arrow opposite $m$ is 1, and the arrow opposite $m'$ is pullback of 1 by part b. – ziggurism Nov 27 '15 at 3:26
• Btw. I asked a semi-related question here here. – Nikolaj-K Nov 27 '15 at 3:27
• Thanks guys, I think I get it. :) – StudentType Nov 27 '15 at 5:19