Is the counit of geometric realization a Serre fibration?

Question

Recall that a Serre fibration between topological spaces is a map which has the homotopy lifting property (HLP) for all CW complexes (equivalently for all disks $D^k$). The Serre fibrations are the fibrations in a model category structure on topological spaces in which the weak equivalences are the weak homotopy equivalences. The cofibrant objects include the CW-complexes.

The geometric realization and singular simplicial set functors from an adjunction $$|-|: sSet \leftrightarrows Top: Sing_*$$ which is in fact a Quillen equivalence between the above model structure and the standard model structure on simplicial sets.

The geometric realization of every simplicial set is a CW-complex and the counit map $$\epsilon_X: |Sing(X)| \to X$$ is a weak homotopy equivalence. Hence this gives a nice functorial CW-approximation for every space. Moreover $|Sing(X)|$, being a CW-complex, is cofibrant.

I would like this to be a cofibrant replacement of $X$, however for that I should further require that the counit map $\epsilon_X$ is also a Serre fibration. This is not so obvious to me because to check the lifting property I have to map into a geometric realization (which is a bit subtle). However using cellular approximation I have been able to show that this map has a weaker fibration property; it is a sort of "Serre fibration version" of a Dold fibration (where we only have a weak homotopy lifting property). This is good enough for many applications, but it still leads me to ask my question:

Is the counit map $\epsilon_X$ a Serre fibration? Are there conditions on X (such as paracompactness) which will make this hold true? If it is not a Serre fibration generally, what is the easiest counterexample?

Another related question is whether in Top there is a functorial cofibrant replacement by CW-complexes. The usual approach would probably be to use the small object argument, which would result in a cellular space (a space like a CW-complex but with the cells possibly attached out of order). I suspect that a little clever handicrafting can make this yield an actual CW-complex. Is that intuition correct?

Answer 1

$\newcommand{\real}[1]{\left\lvert #1 \right\rvert}$$\newcommand{\Sing}[1]{\operatorname{Sing}(#1)}$$\newcommand{\counit}{\epsilon}$$\newcommand{\To}{\longrightarrow}$$\newcommand{\proj}{\mathrm{proj}}$$\newcommand{\NN}{\mathbb{N}}$$\newcommand{\RR}{\mathbb{R}}$Yes, the map $\real{\Sing{X}} \to X$ is a Serre fibration.

[Disclaimer: This answer is very long. A lot of what I will write is contained in Oscar's answer and in the comments. I present it here for completeness and convenience.]

The present answer adapts the constructions given by Oscar Randall-Williams in his answer. The missing point is to prove that the maps generalizing the ones appearing in Oscar's answer are always inclusions of retracts. We will actually prove that they are trivial cofibrations, which will fundamentally require the fact that finite cell complexes are Euclidean neighbourhood retracts. Please upvote Oscar's answer.

Setup

Let $X$ be a topological space, and let $\counit_X:\real{\Sing{X}}\to X$ be the counit map of the adjunction between the singular complex functor and the geometric realization functor. We want to show that $\counit_X$ is a Serre fibration. Let then $h:\real{\Delta^n} \to \real{\Sing{X}}$ and $H:\real{\Delta^n} \times I = \real{\Delta^n \times \Delta^1}\to X$ be continuous maps. We need only prove that it is possible to provide a diagonal lift for the diagram $$ \begin{array}{ccc} \real{\Delta^n} & \overset{h}{\To} & \real{\Sing{X}} \\ \Big\downarrow\rlap{\scriptstyle \iota_0} & & \Big\downarrow\rlap{\scriptstyle \counit_X} \\ \real{\Delta^n}\times I & \underset{H}{\To} & X \end{array} $$

Constructions (adapted from Oscar Randall-Williams' answer)

The space $C$. Since simplices are compact, the image of $h:\real{\Delta^n} \to \real{\Sing{X}}$ is contained in the realization $\real{K}$ of some finite sub-simplicial set $K$ of $\Sing{X}$. Then $h$ restricts to a map $h:\real{\Delta^n}\to\real{K}$, and we can take the mapping cylinder of $h$: $$ C = M_h = \real{K} \coprod_{\real{\Delta^n}} \bigl( \real{\Delta^n}\times I \bigr) $$ This mapping cylinder generalizes the space $C$ described by Oscar Randall-Williams in his answer.

The space $D$. The preceding space $C$ includes naturally into the mapping cylinder $$ D = M_{\proj_{\real{K}}} = \real{K} \coprod_{\real{K}\times\real{\Delta^n}} \bigl( \real{K}\times\real{\Delta^n}\times I \bigr) $$ of the projection map $\proj_{\real{K}} : \real{K}\times\real{\Delta^n} \to \real{K} $. Importantly, observe that since geometric realization preserves colimits and finite products, the space $D$ is also the geometric realization $$ D = \real{M_{\proj_K}} $$ of the simplicial mapping cylinder $M_{\proj_K} = K \coprod_{K\times\Delta^n} (K\times\Delta^n\times\Delta^1)$ of the projection $\proj_K: K\times\Delta^n \to K$.

The space $D$ plays here the role of the join appearing in Oscar Randall-Williams' answer: note that the join $\real{K}\ast\real{\Delta^n}$ is naturally a quotient of $D$.

The maps. The inclusion map $j:C \to D$ is given by:

• $j$ restricted to $\real{K}$ is the canonical inclusion $\real{K}\hookrightarrow D$ of the end of the mapping cylinder;
• $j([x,t]) = [h(x),x,t]$ for $(x,t)\in \real{\Delta^n} \times I$ (where we see $D$ as a quotient of $\real{K}\times\real{\Delta^n}\times I$).

There is a further map $G:C\to X$ determined by:

• $G$ restricted to $\real{K}$ coincides with $\counit_X$;
• $G$ restricted to $\real{\Delta^n}\times I$ coincides with $H$.

Main argument: $j$ is a trivial Hurewicz cofibration

Now that we have adapted Oscar's construction, the main part of the argument consists of showing that $j: C\to D$ is a trivial cofibration.

First, the map $j$ is easily seen to be injective. Since $C$ is compact (because both $\real{K}$ and $\real{\Delta^n}\times I$ are compact) and $D = \real{M_{\proj_K}}$ is Hausdorff, it follows that $j: C\to D$ is a closed map, and in particular a homeomorphism onto its image.

Second, the map $j$ is a homotopy equivalence. Simply note that the composition of $\real{K} \hookrightarrow C \overset{j}{\to} D$ is the canonical inclusion $\real{K} \hookrightarrow M_{\proj_{\real{K}}} = D$ into the mapping cylinder, and thus a homotopy equivalence. Similarly, $\real{K}\hookrightarrow C$ is the inclusion into the mapping cylinder, and thus a homotopy equivalence. By the two-out-of-three property for homotopy equivalences, $j$ is itself a homotopy equivalence.

It remains to show that the inclusion of $j(C)$ into $D$ is a Hurewicz cofibration. Observe that both $C$ and $D$ are finite cell complexes, and in particular are Euclidean neighbourhood retracts (ENRs). This can be proved in the same way as corollary A.10 in the appendix to Allen Hatcher's book "Algebraic topology", which states that finite CW-complexes are ENRs. The desired result is now encoded in the following lemma.

Lemma: Assume $Y$ is a closed subspace of the topological space $Z$. Assume also that $Y$ and $Z$ are both ENRs. Then the inclusion of $Y$ into $Z$ is a Hurewicz cofibration.

This result actually holds in general for ANRs, and is stated as proposition A.6.7 in the appendix of the book "Cellular structures in topology" by Fritsch and Piccinini. Nevertheless, for completeness, I will provide a proof of the lemma at the end of this answer which uses only the results in the appendix of Hatcher's book when $Z$ is compact.

I will now conclude the proof that $\counit_X$ is a Serre fibration.

Conclusion

First, observe that $j:C\to D$ is the inclusion of a strong deformation retract because it is a homotopy equivalence and a Hurewicz cofibration. In particular, $j$ admits a right inverse $r:D\to C$. The composite $\overline{G} = G\circ r:D\to X$ is then an extension of $G:C\to X$ along $j$, i.e. $\overline{G}\circ j = G$.

Now we use the description of $D$ as the realization of the simplicial set $M_{\proj_K}$ to give a diagonal lift for the square diagram at the beginning of this answer. Note that to give a map $\overline{G}:D=\real{M_{proj_K}} \to X$ is by adjunction the same as giving a map $F:M_{proj_K}\to\Sing{X}$. I claim that the composite $$ \widetilde{H} : \real{\Delta^n}\times I \hookrightarrow C \overset{j}{\To} D \overset{\real{F}}{\To} \real{\Sing{X}} $$ is such a diagonal lift:

1. $\counit_X \circ\real{F}= \overline{G}$ by construction of $F$ via adjunction. Consequently, $\counit_X \circ \real{F}\circ j = G$.

2. In particular, $\counit_X \circ \real{(F|_K)} = \counit_X \circ (\real{F})|_{\real{K}} = \counit_X \circ \real{F} \circ j|_{\real{K}} = G|_{\real{K}} = \counit_X$. This implies by adjunction that $F|_K$ is the inclusion of $K$ into $\Sing{X}$. Therefore, the restriction $(\real{F})|_{\real{K}} = \real{(F|_K)}$ is the inclusion of $\real{K}$ into $\real{\Sing{X}}$.

3. It follows that $\widetilde{H}\circ \iota_0 = \real{F}\circ j|_{\real{K}} \circ h = (\real{F})|_{\real{K}} \circ h = h$, i.e. the map $\widetilde{H}$ makes the upper triangle commute.

4. Furthermore, it follows from item 1 that $\counit_X \circ\widetilde{H} = \counit_X \circ\real{F}\circ j|_{(\real{\Delta^n}\times I)} = G|_{(\real{\Delta^n}\times I)} = H$. So $\widetilde{H}$ makes the lower triangle commute.

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Proof of the lemma

We will use (a simplification of) the characterization of Hurewicz cofibrations given in theorem 2 of Arne Strøm's article "Note on cofibrations" (published in Mathematica Scandinavica 19 (1966), pages 11-14).

The inclusion of a closed subspace $Y$ of a metrizable topological space $Z$ is a cofibration if there exists a neighbourhood $U$ of $Y$ in $Z$ which deforms in $Z$ to $Y$ rel $Y$. More explicitly, there must exist a homotopy $F:U\times I\to Z$ such that $F(x,0)=x$ and $F(x,1)\in Y$ for $x\in U$, and $F(y,t)=y$ for $(y,t)\in Y\times I$.

This characterization of closed cofibrations is very similar to (and follows easily from) the usual characterization in terms of NDR-pairs, except that it does not demand the homotopy to be defined on the whole space.

Assuming that $Y$ and $Z$ are ENRs, we will now prove the existence of the neighbourhood $U$ and the homotopy $F$ as above.

A space $A$ is a ENR exactly when:

• $A$ is homeomorphic to a closed subspace of $\RR^N$ for some $N\in\NN$;
• for any $N\in\NN$, if $B$ is a closed subspace of $\RR^N$ which is homeomorphic to $A$, then some neighbourhood of $B$ in $\RR^N$ retracts onto $B$.

[For reference, the aforementioned appendix of Allen Hatcher's book "Algebraic topology" explains these two points when $A$ is compact.]

So we may assume without loss of generality that $Z$ is a closed subspace of $\RR^N$, and that some neighbourhood $V$ of $Z$ in $\RR^N$ retracts to $Z$ via a retraction $r_Z:V\to Z$. Since $Y\subset Z\subset\RR^N$ is closed in $\RR^N$ and $Y$ is a ENR, there also exists a neighbourhood $W$ of $Y$ in $\RR^N$ which admits a retraction $r_Y:W\to Y$. Using the convexity of $\RR^N$, we can produce a straight line homotopy $SLH:W\times I\to\RR^N$ between the identity of $W$ and $r_Y$. We may assume without loss of generality (by shrinking $W$ if necessary) that the image of the homotopy $SLH$ is contained in the open $V$. Define then the desired neighbourhood by $U=W\cap Z$ and the homotopy by $F = r_Z \circ SLH$.

Answer 2

It has path lifting, at least.

Let $(\sigma, t) \in Sing_p(X) \times \Delta^p$, which represents a point in $\vert Sing_\bullet(X) \vert$, and let $\gamma : [0,1] \to X$ be a path starting at $x_0 = \sigma(t)$. We can embed $C := \Delta^p \cup_{t, \{0\}} [0,1]$ into $\Delta^{p+1} = \Delta^p * \{1\}$, and extend the map $\sigma \cup \gamma : C \to X$ to a map $\tau : \Delta^{p+1} \to X$ using that $C \to \Delta^{p+1}$ is a deformation retract. It is then clear that there is a path $\gamma'$ in $\Delta^{p+1}$ so that $(\tau, \gamma'(t))$ maps down to $\gamma(t)$.

This seems fairly natural, so maybe it works in families too?

Answer 3

I think that this is a key test case. Let $u:\Delta^1\to\Delta^2$ be a surjective continuous map, which we can build using a space-filling curve. Let $X$ be the mapping cylinder of $u$, so we have maps $h:[0,1]\times\Delta^1\to X$ and $p:\Delta^2\to X$ with $h(0,x)=pu(x)$. We can regard $p$ as a point in $S_2X$ so it gives a map $p_*:\Delta^2\to |SX|$ with $\epsilon\circ p_*=p:\Delta^2\to X$. If $\epsilon:|SX|\to X$ is a Serre fibration, there must be a map $k:[0,1]\times\Delta^1\to |SX|$ with $k(0,x)=p_*(u(x))$ and $\epsilon(k(t,x))=h(t,x)$. I suspect that this is impossible, but I do not yet have a complete proof.

UPDATE:

OK, here's an attempt to summarise what we learn from the comments. Consider a space $X$, a map $f:\Delta^n\to |SX|$ and a homotopy $h:I\times\Delta^n\to X$ starting with $\epsilon\circ f$. Suppose that f factors through a single cell of $|SX|$. In more detail, we suppose that there is a map $p:\Delta^m\to X$ (which gives rise to $p_*:\Delta^m\to |SX|$) and a map $f_1:\Delta^n\to\Delta^m$ such that $f=p_*\circ f_1$. I think we have an argument that $h$ can be lifted in this case.

Indeed, we can let $C=\Delta^m\cup I\times\Delta^n$ denote the mapping cylinder of $f_1$. After noting that $h(0,x)=\epsilon(f(x))=p(f_1(x))$ we see that $p$ and $h$ fit together to give a map $g:C\to X$. We can also define a continuous injection $j:C\to\Delta^{m+n+1}$ by $j(w)=(w,0)$ on $\Delta^m$ and $j(t,x)=((1-t)f_1(x),tx)$ on $I\times\Delta^n$. This is a closed embedding, because $C$ is compact and $\Delta^{m+n+1}$ is Hausdorff. In particular, this shows that $C$ has finite covering dimension. I think it is not hard to check directly that $C$ is contractible and locally contractible. By results that Ricardo quotes, $C$ is an ANR. As it is also contractible, we should get a retraction $r:\Delta^{m+n+1}\to C$ with $rj=1$. Put $q=gr:\Delta^{m+n+1}\to X$. Define $i:\Delta^m\to \Delta^{m+n+1}$ by $i(w)=(w,0)$ so $qi=p:\Delta^m\to X$. As $i$ is a simplicial structure map it follows that $q_*i=p_*:\Delta^m\to |SX|$. Now define $k:I\times\Delta^n\to |SX|$ by $k(t,x)=q_*((1-t)f_1(x),tx)$, so $$ \epsilon(k(t,x))=q((1-t)f_1(x),tx)=grj(t,x)=g(t,x)=h(t,x). $$ This gives the required homotopy lifting.

FURTHER UPDATE:

I'll leave the above in place for the record, but of course it is superseded by Ricardo's answer. Unfortunately I left the page open and unrefreshed for a long time and did not notice that he was working simultaneously.