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In Strøm's (no relation) paper "The Homotopy Category is a Homotopy Category" he proves that the category of unpointed topological spaces, with Hurewicz fibrations and ordinary cofibrations and homotopy equivalences, is a model category.

Then he has a fairly long section on the pointed case, and his results are not as good: he restricts to well-pointed spaces, and doesn't get all the model category axioms.

But if $\mathcal{M}$ is a model category, then for any object $A\in \mathcal{M}$, the category $A \downarrow \mathcal{M}$ inherits a model category structure in perfectly straightforward way. Since $\mathbf{Top}_* = * \downarrow \mathbf{Top}$, we get a model structure right away.

So: what is going on here? Did he simply miss an easy extension to the pointed case? Is it possible that the natural model structure on $* \downarrow \mathbf{Top}$ is not the one he wants (i.e., the received cofibrations, fibrations or weak equivalences differ somehow from pointed cofibrations, fibrations and pointed homotopy equivalences)?

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    $\begingroup$ Interesting! For what its worth, the fact that $A \downarrow M$ is a model category when $M$ is does not get a lot of emphasis in Quillen's book, which is Str$\emptyset$m's only reference on model categories. In fact, Quillen never clearly states it, as far as I can see: the closest he comes is in section II.2.8, where it is shown that if $M$ is a simplicial closed model category, then so is $M\downarrow X$. Maybe Str$\emptyset$m just missed it? $\endgroup$
    – Charles Rezk
    Commented Nov 30, 2010 at 5:02

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You must allow the weak equivalences to be unpointed homotopy equivalences. These become honest pointed homotopy equivalences between fibrant-cofibrant objects by the generalized whitehead theorem. Strøm's mistake, I think, was that he didn't realize that unbased $Top$ with his model structure has the very special property that all objects are fibrant-cofibrant. This is not the case for based spaces, however (all objects are fibrant but not cofibrant). The cofibrant objects are precisely the nondegenerately based spaces.

By restricting to nondegenerately based spaces, he restricted himself to working in the category of cofibrant objects of the actual model category $*\downarrow Top$ equipped with the relative Strøm model structure. It shouldn't be too surprising that this subcategory does not admit a (compatible) model structure.

(Proof that all objects in Strøm's model structure on $Top$ are fibrant-cofibrant):

Lemma: A Hurewicz fibration $A\to B$ has the RLP with respect to every inclusion of a space $X$ into its cylinder $X\times I$. When $B$ is the terminal object, let $X\times I\to X$ be the map killing the cylinder. Then we get a lift to $X\times I$ of any map $X\to A$ by composing $X\times I\to X\to A$.

Lemma: A closed Hurewicz cofibration is a closed inclusion $A\hookrightarrow B$ such that $A\times I \coprod_{A\times \{0\}} B\times \{0\}\hookrightarrow B\times I$ has the LLP with respect to any map $Y\to *$. When $A$ is empty, this reduces to finding an extension of a map $B\to Y$ to a homotopy extending this, but again, this is immediately possible by composition with the trivial homotopy $B\times I\to B$.

The theorem of Strøm is that there exists a model structure on Top with

  • $C$ = closed hurewicz cofibrations
  • $W$ = homotopy equivalences
  • $F$ = hurewicz fibrations

From which it follows that all objects are fibrant-cofibrant.

(Source of definitions: Dwyer-Spalinski example 3.6)

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  • $\begingroup$ I only included the proof sketch above because David's comment below made me go back and check it. $\endgroup$
    – Harry Gindi
    Commented Dec 1, 2010 at 9:43
  • $\begingroup$ What is the "generalized Whitehead Theorem" you are referring to? Do you have a nice reference for that? $\endgroup$
    – Sebastian Goette
    Commented Jan 25, 2020 at 9:20
  • $\begingroup$ And are there not well-pointed spaces that are homotopy equivalent, but not pointed homotopy equivalent to cofibrant objects? In other words, is the "correct" homotopy category equivalent to the "naive" one, where one takes all objects, and morphisms modulo pointed homotopy? $\endgroup$
    – Sebastian Goette
    Commented Jan 25, 2020 at 9:30
  • $\begingroup$ @SebastianGoette The statement of the generalized Whitehead theorem is essentially this: If $X$ is cofibrant and $Y$ is fibrant, then $C(X,Y)/\sim=\operatorname{Ho}(C)(X,Y)$. Here the relation $\sim$ means homotopy in the model category theoretic sense, which coincides in this particular case with based homotopy of maps. $\endgroup$
    – Harry Gindi
    Commented Jan 25, 2020 at 9:36
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    $\begingroup$ @SebastianGoette I don't know of any counterexamples, but what I wrote in my answer ten years ago was that there is no reason to expect this category to carry such a model category structure. The fact that Strøm's theorem works in the unbased setting is extremely surprising. The original question notes that Strøm was unable to replicate his result in the pointed case. I haven't read his paper, but I would guess that is a nice place to look for explicit counterexamples, if they exist. $\endgroup$
    – Harry Gindi
    Commented Jan 25, 2020 at 10:06
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The weak equivalences in the model structure that $* \downarrow \mathbf{Top}$ inherits formally are pointed maps $f: X\to Y$ that are unpointed homotopy equivalences; but in his discussion of the pointed case, Strøm takes $\mathbf{Top}_*$ to have pointed homotopy equivalences as weak equivalences.

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  • $\begingroup$ @Harry - In what model structure on Top are all objects fibrant and cofibrant?? $\endgroup$
    – David Roberts ♦
    Commented Dec 1, 2010 at 4:27
  • $\begingroup$ @David: Isn't that true in the Strom model structure? $\endgroup$
    – Harry Gindi
    Commented Dec 1, 2010 at 4:43
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    $\begingroup$ Yeah, every object is cofibrant, since the closed HEP for the empty set reduces to a diagram for which we can always find a lift (in particular, $\emptyset \times I=\emptyset$, so we can lift the homotopy however we want (contract $X\times I$ back to $X\times 0$ and compose with the section). Similarly, every object is fibrant for the same trivial reason (see Dywer-Spalinski). $\endgroup$
    – Harry Gindi
    Commented Dec 1, 2010 at 4:56
  • $\begingroup$ Ah, yes. Silly me. $\endgroup$
    – David Roberts ♦
    Commented Dec 1, 2010 at 6:19

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