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1. • CommentRowNumber1.
   • CommentAuthorAndrew Stacey
   • CommentTimeMar 18th 2014
   • PermaLink
   Author: Andrew Stacey
   Format: MarkdownItexHi, I happened to notice today that there are problems with the proof of Lemma 1. Unfortunately, I do not have time to fix this myself, but I thought I'd let you know. It can be fixed as follows. 1) Define $\hat{X}$ to be the pullback of $f \times id : X \times Y \rightarrow Y \times Y$ and the map $Y^{I} \rightarrow Y \times Y$ which is currently denoted (I would not use this notation myself!) by $(d_0,d_1)$. 2) The required map $p : Z \rightarrow Y$ is the composite of the map $Z \rightarrow X \times Y$ which is part of the pullback of 1), and the projection map $X \times Y \rightarrow Y$. 3) The required fibration $Z \rightarrow X$ arises in the same way as in 2), but composing with the the other projection map instead, 4) The required map $X \rightarrow Z$ arises via the universal property of the pullback by using the map $f \circ c : X \rightarrow Y^I$, where $c$ is the map $Y \rightarrow Y^I$ appearing in the factorisation which defines $Y^I$, and the map $id \times f : X \rightarrow X \times Y$.

   Hi,

   I happened to notice today that there are problems with the proof of Lemma 1. Unfortunately, I do not have time to fix this myself, but I thought I’d let you know.

   It can be fixed as follows.

   1) Define $\hat{X}$ to be the pullback of $f \times id : X \times Y \rightarrow Y \times Y$ and the map $Y^{I} \rightarrow Y \times Y$ which is currently denoted (I would not use this notation myself!) by $(d_0,d_1)$.

   2) The required map $p : Z \rightarrow Y$ is the composite of the map $Z \rightarrow X \times Y$ which is part of the pullback of 1), and the projection map $X \times Y \rightarrow Y$.

   3) The required fibration $Z \rightarrow X$ arises in the same way as in 2), but composing with the the other projection map instead,

   4) The required map $X \rightarrow Z$ arises via the universal property of the pullback by using the map $f \circ c : X \rightarrow Y^I$, where $c$ is the map $Y \rightarrow Y^I$ appearing in the factorisation which defines $Y^I$, and the map $id \times f : X \rightarrow X \times Y$.

2. • CommentRowNumber2.
   • CommentAuthorAndrew Stacey
   • CommentTimeMar 18th 2014
   • PermaLink
   Author: Andrew Stacey
   Format: MarkdownItexOkay, so Chrome messes up the preview but is fine once the post is posted. Probably due to the interaction of javascripts between inserting the preview and re-running MathJaX on the page.

   Okay, so Chrome messes up the preview but is fine once the post is posted. Probably due to the interaction of javascripts between inserting the preview and re-running MathJaX on the page.

3. • CommentRowNumber3.
   • CommentAuthorAndrew Stacey
   • CommentTimeMar 18th 2014
   • PermaLink
   Author: Andrew Stacey
   Format: MarkdownItexFor my own future reference, the solution would appear to be [documented here](http://docs.mathjax.org/en/latest/typeset.html).

   For my own future reference, the solution would appear to be documented here.

4. • CommentRowNumber4.
   • CommentAuthorAndrew Stacey
   • CommentTimeMar 18th 2014
   • PermaLink
   Author: Andrew Stacey
   Format: MarkdownItexAlso, the initial post in this thread is a copy of a post by Richard Williamson who reported the issue (so the "I" is not me but him!).

   Also, the initial post in this thread is a copy of a post by Richard Williamson who reported the issue (so the “I” is not me but him!).