## very very elementary lemma on pullbacks which I just cannot prove [closed]

I am new in category theory. I am trying to prove the well known fact that if you have a commutative diagram of the form □□, where each square is a pullback, then the whole diagram is a pullback too, and hence deduce that the pullback of a pullback square is a pullback. Every book I have looked at has this as an exercise, but I (embarrasingly, I know) cannot see the solution. I have tried using the universality property of the two pullbacks but i am lost in calculations. If someone could help, I would really appreciate it.

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Dear Christopher, this question is more suitable for math.stackexchange.com – Fernando Muro May 30 at 13:30
This is something that really needs to be discussed in person, so that your interlocutor can get a better sense of exactly what your difficulties are. – Charles Staats May 30 at 13:31
Crossposted (though with one very less in the title): math.stackexchange.com/questions/151585/… – Rasmus May 30 at 13:47
This is one of those cases where it's easier to write the pullback algebraically rather than thinking about it pictorially, since the algebra gives you a cancellation rule. Here's the proof: $P = Y\times_X Z = Y\times X (X\times_A B) = Y\times_A B$, where $P$ is the upper left corner, the left square uses $X$, $Y$, and $Z$, and $Z$ is the upper left of the right square, i.e. is the pullback of $X$ and $B$ over $A$ (the lower right corner) – David White May 30 at 15:53
@David: This argument is circular. – Martin Brandenburg May 30 at 17:12
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