5
$\begingroup$

Let $A$ be a topological space, and $S(A)$ its singular simplicial set. If $|S(A)|$ denotes the geometric realisation, it is known that the co-unit $\epsilon_A\colon|S(A)|\to A$ is a weak-equivalence.

Now, since the functor $S$ preserves weak-equivalences, the map $S(\epsilon_A)\colon S(|S(A)|)\to S(A)$ is still a weak equivalence. Since both simplicial sets are Kan complexes, it is in fact an homotopy equivalence, and one inverse is given by the unit of the adjunction $\eta_{S(A)}\colon S(A)\to S(|S(A)|)$.

One checks easily by hand that $S(\epsilon_A)\circ\eta_{S(A)}=\text{Id}_{S(A)}$. The question is, does there exist an explicit homotopy between $\eta_{S(A)}\circ S(\epsilon_A)$ and $\text{Id}_{S(|S(A)|)}$, preferably functorial in $A$?

Edit : after a bit of work, I reformulated my question into several questions that do not rely on any explicit homotopy. I am still interested in an answer for the previous question, but any answer to the reformulated ones would be helpfull.

(1) : Let $K$ be a simplicial set (one might assume that it is the nerve of a (finite) poset if needed). Taking the realisation, and applying the previous construction, we have an homotopy equivalence given by $S(\epsilon_{|K|})\colon S(|S(|K|)|)\to S(|K|)$ and $\eta_{S(|K|)}\colon S(|K|)\to S(|S(|K|)|)$. We have $S(\epsilon_{|K|})\circ\eta_{S(|K|)}=\text{Id}_{S(|K|)}$, the first question is, does there exist an homotopy $H\colon S(|S(|K|)|)\times \Delta^1\to S(|S(|K|)|)$ between $\eta_{S(|K|)}\circ S(\epsilon_{|K|})$ and $\text{Id}_{S(|S(|K|)|)}$ such that the following diagram commutes

$$\require{AMScd}\begin{CD} K\times\Delta^1@>\text{pr}_K>>K\\ @V(\eta_{S(|K|)}\circ\eta_K)\times \Delta^1 VV@VV\eta_{S(|K|)}\circ\eta_K V\\ S(|S(|K|)|)\times \Delta^1@>H>>S(|S(|K|)|) \end{CD}$$

(2) Let $f\colon A\to |K|$ be a continous map, where $K$ is the same simplicial set as in (1), and let $H$ be the homotopy obtained in (1) (or any fixed homotopy). Does there exist a homotopy $H'\colon S(|S(A)|)\times \Delta^1\to S(|S(A)|)$ between $\eta_{S(A)}\circ S(\epsilon_A)$ and $\text{Id}_{S(|S(A)|)}$ such that the following diagram commutes : $$\begin{CD} S(|S(A)|)\times \Delta^1@>H'>>S(|S(A)|)\\ @VS(|S(f)|)\times \Delta^1 VV@VVS(|S(f)|)V\\ S(|S(|K|)|)\times \Delta^1@>H>>S(|S(|K|)|) \end{CD}$$

Edit 2: The answer to (1) seems to be positive. Indeed in Piccinini and Fritsch Cellular structures in topology, in Proposition 4.5.29, they show that for any topological space $T$, there is an homotopy relative to $S(T)$ between $\text{Id}$ and $\eta_{S(T)}\circ S(\epsilon_T)$, which is stronger than (1).

I still can't figure out the answer to (2), even though I suspect that the proofs of Proposition 4.5.29 and the preceeding theorem could be adapted to provide such a homotopy. Any help on this point would be appreciated.

Fritsch, Rudolf; Piccinini, Renzo A., Cellular structures in topology, Cambridge Studies in Advanced Mathematics. 19. Cambridge etc.: Cambridge University Press. xi, 326 p. (1990). ZBL0837.55001.

$\endgroup$
  • $\begingroup$ This is just a coincidence. I was reading something about homotopy for functors from ams.org/journals/proc/1972-036-02/S0002-9939-1972-0334212-5/…. There is some discussion about this. You might find that article useful.. My question is mathoverflow.net/questions/301339/homotopy-for-functors $\endgroup$ – Praphulla Koushik May 28 '18 at 15:12
  • $\begingroup$ The question at mathoverflow.net/questions/171662 is in some sense adjoint to this one, so the formulae there might suggest a useful approach. $\endgroup$ – Neil Strickland May 31 '18 at 15:43
  • 2
    $\begingroup$ I was aware of this question when posting mine. Since I am working with simplicial sets and not simplicial spaces, the map $|S(A)|\to A$ might only be a weak equivalence. Nevertheless, I tried adapting this homotopy to $S(|S(A))\to S(|S(A)|)$ with no success. The main problem comes from the fact that the map $A\to |S(A)|$ sending $a\in A$ to the corresponding $0$-simplex in $|S(A)|$ is almost never continuous. $\endgroup$ – S. Douteau Jun 1 '18 at 11:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.