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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeFeb 10th 2015
   • (edited Feb 10th 2015)
   • PermaLink
   Author: Urs
   Format: MarkdownItexIn the context of the discussion at [[infinity-action]], given some $X$ and some $A$, consider $[X,A]$ as equipped with the canonical $Aut(X)$-action. Given any $f \colon X \to A$ we have a homotopy pullback of the form $$ \array{ Aut(X) &\longrightarrow& \ast \\ \downarrow && \downarrow^{\mathrlap{f}} \\ [X,A] &\longrightarrow& [X,A]//Aut(X) } $$ I am sure that the left vertical map here is the one known as $f \circ (-)$. But I am being dense and lacking a real formal proof. What's a good formal proof of this?

   In the context of the discussion at infinity-action, given some $X$ and some $A$, consider $[X,A]$ as equipped with the canonical $Aut(X)$-action. Given any $f \colon X \to A$ we have a homotopy pullback of the form

   $$\array{ Aut(X) &\longrightarrow& \ast \\ \downarrow && \downarrow^{\mathrlap{f}} \\ [X,A] &\longrightarrow& [X,A]//Aut(X) }$$

   I am sure that the left vertical map here is the one known as $f \circ (-)$. But I am being dense and lacking a real formal proof. What’s a good formal proof of this?

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeFeb 10th 2015
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   Author: Mike Shulman
   Format: MarkdownItexWell, the pullback square exhibiting $[X,A]$ as the fiber of $[X,A]\sslash Aut(X) \to \mathbf{B} Aut(X)$ induces this map by the universal property...

   Well, the pullback square exhibiting $[X,A]$ as the fiber of $[X,A]\sslash Aut(X) \to \mathbf{B} Aut(X)$ induces this map by the universal property…

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeFeb 10th 2015
   • PermaLink
   Author: Urs
   Format: MarkdownItexHow do you see this? I am talking about identifying the universal map as the one given by restricting to equivalences the map $[X,X] \to [X,A]$ formed by actual composition. Is this obvious? What am I missing?

   How do you see this?

   I am talking about identifying the universal map as the one given by restricting to equivalences the map $[X,X] \to [X,A]$ formed by actual composition.

   Is this obvious? What am I missing?

4. • CommentRowNumber4.
   • CommentAuthorMike Shulman
   • CommentTimeFeb 10th 2015
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   Author: Mike Shulman
   Format: MarkdownItexI think it should follow by unraveling the universal property and using the fact that the equivalence $X\simeq X$ corresponding to a point of $Aut(X)$ is also the path living in the homotopy pullback square exhibiting $Aut(X)$ as $\Omega \mathbf{B} Aut(X)$. I don't have time to write it out carefully right now.

   I think it should follow by unraveling the universal property and using the fact that the equivalence $X\simeq X$ corresponding to a point of $Aut(X)$ is also the path living in the homotopy pullback square exhibiting $Aut(X)$ as $\Omega \mathbf{B} Aut(X)$. I don’t have time to write it out carefully right now.

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeFeb 11th 2015
   • (edited Feb 11th 2015)
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   Author: Urs
   Format: MarkdownItexSo that's why I said I am sure that it works, but that I don't have a fully formal proof. Sorry, I didn't mean to ask you to do this work for me. Though I thought I'd try and see if you see this within microseconds. :-)

   So that’s why I said I am sure that it works, but that I don’t have a fully formal proof.

   Sorry, I didn’t mean to ask you to do this work for me. Though I thought I’d try and see if you see this within microseconds. :-)

6. • CommentRowNumber6.
   • CommentAuthorMike Shulman
   • CommentTimeFeb 11th 2015
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexSorry. (-:

   Sorry. (-: