Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

• Forgot your password?
• Apply for membership

1. • CommentRowNumber1.
   • CommentAuthorFinnLawler
   • CommentTimeAug 8th 2010
   • PermaLink
   Author: FinnLawler
   Format: MarkdownItexAdded a proof of the pasting lemma to [[pullback]], and the corresponding lemma to [[comma object]] (also added the construction by pullbacks and cotensors there).

   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).

2. • CommentRowNumber2.
   • CommentAuthorMarc
   • CommentTimeAug 10th 2010
   • PermaLink
   Author: Marc
   Format: HtmlActually the statement of your proposition is wrong. If the right square in <pre> A --> B --> C | | | v v v D --> E --> F </pre> 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 <pre> A ==== A ==== A | | | | |i | | v | A ---> B ---> A i p </pre> 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.

   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.
3. • CommentRowNumber3.
   • CommentAuthorFinnLawler
   • CommentTimeAug 10th 2010
   • (edited Aug 10th 2010)
   • PermaLink
   Author: FinnLawler
   Format: MarkdownItexWell 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.

   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.