Timeline for Inner hom and geometric realization.
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 22, 2011 at 23:30 | comment | added | Alan Wilder | Ahh, that is the piece I was missing. Thank you! | |
| Feb 18, 2011 at 19:52 | vote | accept | Alan Wilder | ||
| Feb 18, 2011 at 19:52 | vote | accept | Alan Wilder | ||
| Feb 18, 2011 at 19:52 | |||||
| Feb 18, 2011 at 16:45 | comment | added | Charles Rezk | If $f:Y\to Z$ happens to be a trivial cofibration between Kan complexes, then $Y$ is a "simplicial deformation retract" of $Z$, i.e., you can produce a simplicial homotopy equivalence with $gf=$ identity. | |
| Feb 18, 2011 at 16:44 | comment | added | Charles Rezk | It's not really necessary to know that it is a trivial cofibration. What might be helpful to know is that if $f:Y\to Z$ is a weak equivalence between Kan complexes, then it is a "simplicial homotopy equivalence", i.e., there is a map $g:Z\to Y$ and simplicial homotopies $\Delta[1]\times Y\to Z$ and $\Delta[1]\times Z\to X$ between the composites and the identity. Then it would follow that $\underline{sSet}(X,Y)\to \underline{sSet}(X,Z)$ is a simplicial homotopy equivalence for any $X$. | |
| Feb 18, 2011 at 4:54 | comment | added | Alan Wilder | In particular, based on your answer in my earlier question about classifying space functor in the same setup, you gave an example showing that the target being a groupoid (Kan in this setup) was essential. But the unit map is always a trivial cofibration. | |
| Feb 18, 2011 at 4:48 | comment | added | Alan Wilder | Well, what was confusing me was that it didn't seem like the unit map could possibly be a fibration, since the target is so much "bigger". I should have said that. Why would it being a cofibration help? Is it true that a map $$ \underline{\mathbf{sSET}} (X,Y) \rightarrow \underline{\mathbf{sSET}} (X,Z) $$ induced by a trivial cofibration $Y\rightarrow Z$, and where $X$ is cofibrant is a weak equivalence as well? I thought the map in the covariant factor needed to be a fibration. | |
| Feb 18, 2011 at 3:28 | history | answered | Charles Rezk | CC BY-SA 2.5 |