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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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  • $\begingroup$ See also this question. $\endgroup$
    – t.b.
    Commented May 30, 2012 at 13:41
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    $\begingroup$ Crossposted: mathoverflow.net/questions/98378/… $\endgroup$
    – Rasmus
    Commented May 30, 2012 at 13:49
  • $\begingroup$ 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. $\endgroup$
    – MJD
    Commented May 30, 2012 at 14:18

2 Answers 2

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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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    $\begingroup$ 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? $\endgroup$ Commented May 31, 2012 at 4:24
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    $\begingroup$ That is a very good question. $\endgroup$ Commented May 31, 2012 at 7:16
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    $\begingroup$ Thanks for a citeable reference for this proof! Trivial nitpick: $m = m' = n' \circ f$ on page 2 should be $m = m' = f \circ n'$. $\endgroup$ Commented Jun 30, 2015 at 12:56
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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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