5
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ See also this question. $\endgroup$
    – t.b.
    May 30, 2012 at 13:41
  • 1
    $\begingroup$ Crossposted: mathoverflow.net/questions/98378/… $\endgroup$
    – Rasmus
    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
    May 30, 2012 at 14:18

2 Answers 2

18
$\begingroup$

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

$\endgroup$
3
  • 1
    $\begingroup$ 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? $\endgroup$ May 31, 2012 at 4:24
  • 1
    $\begingroup$ That is a very good question. $\endgroup$ May 31, 2012 at 7:16
  • 2
    $\begingroup$ Thanks for a citeable reference for this proof! Trivial nitpick: $m = m' = n' \circ f$ on page 2 should be $m = m' = f \circ n'$. $\endgroup$ Jun 30, 2015 at 12:56
3
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.