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• CommentRowNumber1.
• CommentAuthorFinnLawler
• CommentTimeAug 8th 2010

Added a proof of the pasting lemma to pullback, and the corresponding lemma to comma object (also added the construction by pullbacks and cotensors there).

• CommentRowNumber2.
• CommentAuthorMarc
• CommentTimeAug 10th 2010
Actually the statement of your proposition is wrong. If the right square in

A --> B --> C
| | |
v v v
D --> E --> F

is a pullback then the two statements

(i) the outer rectangle is a pullback
(ii) the left square is a pullback

are equivalent. But it may well happen that both (i) and (ii) hold,
without the right square being a pullback.

For instance let i: A --> B be a split mono with retraction p: B --> A
and consider

A ==== A ==== A
| | |
| |i |
| v |
A ---> B ---> A
i p

where all unnamed maps are identity maps. Then the left square and
the outer rectangle are pullbacks but the right square cannot be
a pullback unless i was already an isomorphism.
• CommentRowNumber3.
• CommentAuthorFinnLawler
• CommentTimeAug 10th 2010
• (edited Aug 10th 2010)

Well spotted. The mistake in the ’proof’ is in confusing cones over $(d \to e \to f, c \to f)$ and cones over $(d \to e, e \to f, c \to f)$. I think they are the same iff the right square is a pullback.

Edit: I made the same mistake at comma object. Both fixed now, I think.