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I'm trying to solve a problem which asks me to deduce that "the pullback of a pullback square is a pullback", using the result of http://ncatlab.org/nlab/show/pullback (under 'Pasting of pullbacks') that after concatenating 2 commutative squares into a larger rectangle, if the right-hand square is a pullback, the left-hand square is a pullback iff the larger rectangle is a pullback.

So, I gather I'm meant to be forming some sort of cube and using the result to show various combinations of faces make pullbacks, but I'm not sure what it actually means to say "the pullback of a pullback": I guess a pullback square is in a sense 2 commuting morphisms which you could take the pullback of, but when the problem says "the pullback of a pullback", clearly they don't just mean to pull back these 2 commuting morphisms because that's obviously a pullback. I think I'm meant to show that something is in fact that pullback of these 2 commuting morphisms, but what is it? Thanks for the help!

Spyam
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    Take 8 objects, and form a cube, with all arrows directed towards the back-right-bottom corner. Suppose the back, top, bottom, left, and right faces are pullback squares. Show the front is also a pullback square. (You actually only need some of the faces to be pullbacks.) – Zhen Lin Nov 09 '11 at 21:14
  • So what is it you think 'pullback of a pullback square' actually means, sorry? Why it should mean "a cube with all but the front face pullbacks" is not clear to me. (Though I don't doubt that this cube you suggest may be equivalent to the way it is meant to be interpreted, it doesn't seem obvious to me why 'pullback of a pullback square' should be that, but perhaps that's just me being stupid.) – Spyam Nov 09 '11 at 21:31
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    I just restated the exercise more explicitly. I know it's the correct interpretation because the lecturer told me. – Zhen Lin Nov 09 '11 at 21:35
  • @ZhenLin: Please consider converting your comment into an answer, so that it gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Lord_Farin Jun 08 '13 at 13:15

2 Answers2

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This CW answer intends to remove this question from the Unanswered queue.

The intended exercise is, as Zhen Lin points out:

Take 8 objects, and form a cube, with all arrows directed towards the back-right-bottom corner. Suppose the back, top, bottom, left, and right faces are pullback squares. Show the front is also a pullback square. (You actually only need some of the faces to be pullbacks.)

Community
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Lord_Farin
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The general result is the following:

Proposition. Suppose we have a commutative cube in a category: enter image description here

Assume the front and bottom faces are pullback squares. Then the following are equivalent:

  1. The total rectangle from $U'\to V'$ to $X\to Y$ is a pullback.

  2. The top and back faces are pullback squares.

  3. The top face or the back face is a pullback square.

The proof comes from the general pasting law for pullbacks.

Elías Guisado Villalgordo
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