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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeOct 28th 2019
   • (edited Oct 28th 2019)
   • PermaLink
   Author: Urs
   Format: MarkdownItexPlease excuse me for a basic topology question: What's the homotopy type of the homotopy pushout $$S^3 \underset{S^3 \times S^1}{\sqcup} S^1$$ induced by the two projection maps, hence of the ordinary pushout $$S^3 \times \mathbb{D}^2 \underset{S^3 \times S^1}{\sqcup} \mathbb{D}^4 \times S^1$$ induced by the boundary inclusions?

   Please excuse me for a basic topology question:

   What’s the homotopy type of the homotopy pushout

   $$S^3 \underset{S^3 \times S^1}{\sqcup} S^1$$

   induced by the two projection maps,

   hence of the ordinary pushout

   $$S^3 \times \mathbb{D}^2 \underset{S^3 \times S^1}{\sqcup} \mathbb{D}^4 \times S^1$$

   induced by the boundary inclusions?

2. • CommentRowNumber2.
   • CommentAuthorDylan Wilson
   • CommentTimeOct 28th 2019
   • PermaLink
   Author: Dylan Wilson
   Format: TextThis is the join, so it'll be S^5

   This is the join, so it'll be S^5
3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeOct 28th 2019
   • (edited Oct 28th 2019)
   • PermaLink
   Author: Urs
   Format: MarkdownItexAh, right. Thanks! (and our pages on joins need some improvement, too...)

   Ah, right. Thanks!

   (and our pages on joins need some improvement, too…)

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeOct 28th 2019
   • (edited Oct 28th 2019)
   • PermaLink
   Author: Urs
   Format: MarkdownItexLet me try to say something more interesting: I was trying to see if an unordered configuration space of points could be realized as a mapping space. I'll write $Conf(\Sigma, A)$ for the space of un-ordered configurations of points in $\Sigma$ with labels in the pointed space $A$. Now a cyclically ordered configuration in $S^3$ should equivalently be a element in the fiber product $Conf(S^3, S^1) \times_{Conf(S^3\times S^1, \emptyset)} Conf(S^1, S^3)$. [edit: yeah, there is a problem with this fiber product] I was trying to guess that this is essentially equivalent to the maps into $S^4$ out of that join of $S^3$ with $S^1$. But this must be wrong. Hm...

   Let me try to say something more interesting:

   I was trying to see if an unordered configuration space of points could be realized as a mapping space.

   I’ll write $Conf(\Sigma, A)$ for the space of un-ordered configurations of points in $\Sigma$ with labels in the pointed space $A$.

   Now a cyclically ordered configuration in $S^3$ should equivalently be a element in the fiber product $Conf(S^3, S^1) \times_{Conf(S^3\times S^1, \emptyset)} Conf(S^1, S^3)$.

   [edit: yeah, there is a problem with this fiber product]

   I was trying to guess that this is essentially equivalent to the maps into $S^4$ out of that join of $S^3$ with $S^1$. But this must be wrong. Hm…