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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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See also this question. – t.b. May 30 '12 at 13:41
Crossposted:… – Rasmus May 30 '12 at 13:49
You might also try posting your work up to the point where you get lost in calculations; then someone might help you find your way out again. – MJD May 30 '12 at 14:18

Just for you, and it turns out my answer has to contain at least 30 characters, so let's make it 100.

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A question to whoever knows these things -- is it possible to include commutative diagrams in answers here, so that the external PDF isn't necessary? – ShreevatsaR May 31 '12 at 4:24
That is a very good question. – Andrej Bauer May 31 '12 at 7:16
Thanks for a citeable reference for this proof! Trivial nitpick: $m = m' = n' \circ f$ on page 2 should be $m = m' = f \circ n'$. – darij grinberg Jun 30 '15 at 12:56

My advice: try once more. The pullback property for the big square (or rectangle) really follows from the pullback properties of the two small squares.

Hint: Start with the right hand square. Can you see two arrows as in the condition of the pullback property?

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