I have seen it claimed that (for compactly generated Hausdorff spaces) the geometric realization of the singular (internal) simplicial space is homotopy equivalent to the original space. I know how to prove this when the space is sufficiently nice (metric spaces). However I do not understand the proof in full generality, and I have begun wonder if perhaps this is sometimes false for sufficiently exotic spaces?

4 years ago I asked a similar MO question, however that question, while nearly the same, was really after something quite different. In particular, at the time I said I was fine making niceness assumptions about the spaces involved. This question, however is precisely about what happens if you do not make these assumptions.


For this question let us work with compactly generated Hausdorff spaces. "Topological space" will mean one of these. A simplicial space is a functor from the opposite of the combinatorial simplex category $\Delta$ to the category of topological spaces. Simplicial spaces form a category sTop. There is an adjunction:

$$ |-|: sTop \leftrightarrows Top: Sing $$

Where $Sing(X)$ is the simplical space whose n-th space is $map(\Delta^n, X)$ with the compactly generated compact-open topology.

We can also form $cX$, the constant simplicial space, and there is a map $cX \to Sing(X)$ which on each level includes $X$ as the constant maps from $\Delta^n$ to $X$. Upon geometric realization $|cX| \cong X$ (homeomorphism). We have that the composite $$X = |cX| \to |Sing(X)| \to X$$ is the identity map. Moreover the inclusion $cX \to Sing(X)$ is a levelwise homotopy equivalence. All this is clear.

Now the closely related fat geometric realization always sends levelwise homotopy equivalences to homotopy equivalences of spaces. For the question below it is crucial that we are using the standard geometric realization.


Question 1: When is the counit $\epsilon_X: |Sing(X)| \to X$ a homotopy equivalence? a weak homotopy equivalence?

Now the ordinary geometric realization does not always send levelwise homotopy equivalences to homotopy equivalences. In order to send levelwise homotopy equivalences to homotopy equivalences we need to assume a cofibration property. Specifically it is known that levelwise homotopy equivalences between proper simplicial spaces pass to homotopy equivalences of spaces after geometric realization.

So a closely related question is:

Question 2: When is the singular simplicial space proper?

The constant simplicial space $cX$ is always proper, so the counit $\epsilon_X$ will be a homotopy equivalence whenever $Sing(X)$ is proper.

Of course maybe we can get by with less:

Question 3: Is there some property more general than proper which is satisfied by the singular simplicial space, which would allow us to conclude that geometric realization sends levelwise homotopy equivalence to homotopy equivalences? (or weak homotopy equivalences?)

Now on this n-lab page (dated 12 June, 2014) in Proposition 4 it states that the counit map $\epsilon_X$ is always a weak homotopy equivalence. It cites Seymour Prop. 3.1:

R. M. Seymour, Kan fibrations in the category of simplicial spaces Fund. Math., 106(2):141-152, 1980.

Seymour actually claims that it is a homotopy equivalence. At one stage in the proof, he states "Now it is clear that cX and Sing(X) are proper...". However he does not give a proof of this and in fact I believe, after helpful conversations with colleague, that this is actually false. That there are spaces $X$ for which the singular simplicial space $Sing(X)$ is not proper. A counter example will be provide below.

This question was also raised by my colleague Karol Szumiło on the n-forum, but without resolution.

singular simplicial spaces are not always proper

In this example we take $X = I^{A}$ where $A$ is any uncountable set, i.e. $X$ is the cube of uncountable dimension. It is compact and hence compactly generated.

Now for $Sing(X)$ to be proper we need that certain maps built out of it should be closed Hurwicz cofibrations. The first such map is the first degeneracy map:

$$ X \to PX = maps(\Delta^1, X) $$

which realizes $X$ as the constant paths in the free path space. Now to be a closed Hurewicz cofibration it is necessary that there exists a function

$$ u: PX \to [0,1]$$

such that $X = u^{-1}(0)$ is exactly the zero-set of the function. This implies that $X \subseteq PX$ must be the intersection of a countable family of open subsets (the $u^{-1}( [0, \frac{1}{n}) )$ ).

However a simple inspection shows that this is not the case for $X \subseteq PX$. The space $X$ is not the intersection of any countable collection of open subsets of $PX$. (Essentially any countable intersection of opens still allows paths which are free to "wiggle" in at least uncountably many directions).


The singular simplicial space $Sing(X)$ is not proper for $X$ a cube of uncountable dimension.

so that question 2 does not always hold.

  • $\begingroup$ Perhaps I just misunderstand your definition of the geometric realization $\mid . \mid$, but ... isn't a constant map corresponding to a full n-simplex in $cX$? (So points in $X$ are simplices, not just points in $\mid cX\mid$?) If so, what do you mean by $X=\mid cX\mid$? $\endgroup$
    – ThiKu
    Jun 12, 2014 at 12:34
  • $\begingroup$ @user39082: A point $x\in X$ gives an element $c^n_x\in(cX)_n$ for each $n$, and thus points $[c^n_x,u]\in|cX|$ for each $u\in\Delta_n$. However, $c^n_x$ is a degeneration of $c^0_x$, and if you think about how the simplicial identifications work, you will see that $[c^n_x,u]$ depends only on $x$, not $n$ or $u$. $\endgroup$ Jun 12, 2014 at 12:41

1 Answer 1


You can just write down the required homotopy.

A point in $|\text{Sing}(X)|$ is an equivalence class $[\sigma,u]$ where $u\in\Delta_n$ and $\sigma:\Delta_n\to X$. Define $\theta^n_{u,t}:\Delta_n\to\Delta_n$ by $\theta^n_{u,t}(x)=tx+(1-t)u$. Then define $\phi_t[\sigma,u]=[\sigma\circ\theta^n_{u,t},u]$. To see that this is well-defined, suppose that $\sigma=\tau\circ\Delta_\alpha$ for some $\alpha:[n]\to[m]$. Put $v=\Delta_\alpha(u)\in\Delta_m$, so that $[\sigma,u]=[\tau,v]$. As $\Delta_\alpha$ is an affine map, we have $\Delta_\alpha\circ\theta^n_{u,t}=\theta^m_{v,t}\circ\Delta_\alpha$ and so $\sigma\circ\theta^n_{u,t}=\tau\circ\theta^m_{v,t}\circ\Delta_\alpha$, which is what we need. Now $\phi_1$ is clearly the identity, but $\phi_0([\sigma,u])=[c_{\sigma(u)}\circ\Delta_\pi,u]$, where $\pi$ is the unique map $[n]\to[0]$ and $c_{\sigma(u)}:\Delta_0\to X$ has value $\sigma(u)$. Here $[c_{\sigma(u)}\circ\Delta_\pi,u]$ is the same as $[c_{\sigma(u)},\Delta_\pi(u)]=[c_{\sigma(u)},*]$, where $*$ denotes the unique point of $\Delta_0$. In other words, $\phi_0$ is the composite of the obvious maps $|\text{Sing}(X)|\to X\to |\text{Sing}(X)|$.

  • 4
    $\begingroup$ Oh Fantastic! Of course the maps from a simplex into any space X contract onto the constant maps, but not in a canonical way. I hadn't realized the (obvious in retrospect) fact that the extra data of a point in the realization (namely a point in the simplex) gives us a canonical way to make this contraction. Very satisfying, thank you! $\endgroup$ Jun 12, 2014 at 14:38
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    $\begingroup$ I have a clarification to ask. Another MO answer (mathoverflow.net/a/25547/5792) points out that if Sing(X) means the simplicial set, then |Sing(X)| is a CW complex, so it's definitionally hopeless the two be homotopy equivalent if X is not homotopy equivalent to a CW complex. This answer avoids contradicting that one because it uses the realization of a simplicial space rather than that of a simplicial set, which needn't be a CW complex. But to avoid a contradiction it is necessary your answer fail for the realization of the simplicial set |Sing(X)|. ... $\endgroup$
    – jdc
    Jun 3, 2017 at 2:58
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    $\begingroup$ ... Is the point that your $\phi_t$ is discontinuous for the latter because the first coordinate $\sigma \circ \theta^n_{u,t}$ is now on a space with discrete topology? $\endgroup$
    – jdc
    Jun 3, 2017 at 2:59
  • 3
    $\begingroup$ @jdc Yes, that's the point $\endgroup$ Jun 5, 2017 at 1:49
  • $\begingroup$ That $\phi_t$ is pretty bold o.0 ! $\endgroup$ Dec 15, 2020 at 16:54

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