Strøm model structures on the category of simplicial sets

Question

Let $X,Y$ be simplicial sets. A simplicial homotopy is a simplicial map of the form $h:X\times\Delta^1\rightarrow Y$. There are two distinguished maps $$ in_0:X\cong X\times\Delta^0\xrightarrow{1\times d^1}X\times\Delta^1\xleftarrow{1\times d^0}X\times\Delta^0\cong X:in_1 $$ which include in at the $0$ and $1$ vertices, respectively. Given simplicial maps $f,g:X\rightarrow Y$, say that $h$ is a homotopy from $f$ to $g$ and write $h:f\sim g$ if $f=h\circ in_0$ and $g=h\circ in_1$. This defines a reflexive relation on the set of simplicial maps $X\rightarrow Y$ which generally fails to be either symmetric or transitive.

To rectify this defect let $\simeq$ be the equivalence relation generated by $\sim$. Thus simplicial maps $f,g:X\rightarrow Y$ are related by $\simeq$ if there is a sequence $f_0,\dots,f_n:X\rightarrow Y$ such that $f=f_0$, $g=f_n$, and either $f_i\sim f_{i+1}$ or $f_{i+1}\sim f_i$ for each $i=0,1\dots,n-1$.

The relation $\sim$ is compatible with composition, so $\simeq$ defines a congruence on $sSet$. There is an associated quotient category $hsSet$ with hom sets $hsSet(X,Y)=sSet(X,Y)/\simeq$. Say that a simplicial map $f:X\rightarrow Y$ is a simplicial homotopy equivalence if it becomes an isomorphism when projected into $hsSet$, e.g., if there is $g:Y\rightarrow X$ such that $gf\simeq id_X$ and $fg\simeq id_Y$.

Every simplicial homotopy equivalence is a simplicial weak equivalence. On the other hand, simplicial weak equivalences between Kan complexes are simplicial homotopy equivalences. The full subcategory of $hsSet$ spanned by the Kan complexes is exactly the standard homotopy category of $sSet$ in its Quillen structure (up to equivalence).

The collection $$\mathcal{W}_{she}=\{\text{simplicial homotopy equivalences}\}$$ is a saturated class of weak equivalences in $sSet$.

Question: Is there a model structure on $sSet$ with weak equivalences $\mathcal{W}_{she}$?

I'm assuming that this is not true, but I'm looking for an easy (preferably elementary) proof of this.

References:

P. Goerss, J. Jardine, Simplicial Homotopy Theory, Birkhäuser Verlag, (2009).

Answer 1

Let’s prove that no such model structure exists, following Tom Goodwillie’s answer, and comments from Tom and from Tyrone. See Tom's comment for a simplified version of the following argument.

In the following, assume there is a model structure on $sSet$ with weak equivalences the simplicial homotopy equivalences. All model-categorical terms refer to this putative model structure. We will derive a contradiction.

Lemma 1: Let $I$ be a simplicial-homotopy-contractible simplicial set $I$. Then $I$ has at most one cofibrant point.

Proof: Suppose there are two distinct cofibrant points $i_0 \neq i_1$. Then $i_0,i_1 : \Delta^0 \rightrightarrows I$ are trivial cofibrations. And the transfinite composite $\Delta^0 \xrightarrow{i_0} I {}_{i_1} \cup_{i_0} I \to I {}_{i_1} \cup_{i_0} I {}_{i_1} \cup_{i_0} I \to \cdots $ is a trivial cofibration from $\Delta^0$ to a simplicial set which is not simplicial-homotopy contractible — contradiction.

Lemma 2: $\Delta^0$ is cofibrant.

Proof: There exists a nonempty cofibrant object $C$ and so $\Delta^0$ is cofibrant as a retract of $C$.

Lemma 3: The map $\partial \Delta^1 \to \Delta^0$ is a cofibration. (Thus every trivial fibration is injective on vertices, and every trivially fibrant object has a unique vertex.)

Proof: Let $\partial \Delta^1 \to I \to \Delta^0$ be a cofibration / trivial fibration factorization. By Lemma 2, the maps $\Delta^0 \rightrightarrows \partial \Delta^1$ are cofibrations, and so the composite maps $\Delta^0 \rightrightarrows I$ are cofibrations. By Lemma 1, these must be the same map. So $\partial \Delta^1 \to \Delta^0$ is a retract of $\partial \Delta^1 \to I$ and hence is a cofibration.

Lemma 4: The initial endpoint inclusion $i_0 : \Delta^0 \to \Delta^1$ is either a trivial cofibration or else a trivial fibration.

Proof: Let $\Delta^0 \to X \to \Delta^1$ be a (trivial cofibration, trivial fibration) factorization of $i_0$. By pullback, the fiber $X_0 = i_0^{-1}(\{0\}) \subseteq X$ is trivially fibrant. So by Lemma 3, $X_0$ has exactly one vertex $x_0$, the image of $\Delta^0$. Therefore if $X \to \Delta^1$ is surjective, then there is an edge in $X$ emanating from $x_0$ which maps down to the nondegenerate 1-simplex of $\Delta^1$. So $i_0$ is a retract of $\Delta^0 \to X$ and hence a (trivial) cofibration.

Otherwise, $X \to \Delta^1$ is not surjective, and therefore factors through $\partial \Delta^1$. Since $X$ is simplicially homotopy contractible, it is in particular connected, so we must have $X = X_0$. It follows that $i_0$ is a retract of $X \to \Delta^1$ and hence is a trivial fibration.

Proposition: There is no model structure on $sSet$ with weak equivalences the simplicial homotopy equivalences.

Proof: By Lemma 4, the endpoint inclusion $i_0 : \Delta^0 \to \Delta^1$ is either a trivial cofibration or a trivial fibration. In the first case, Tom’s argument (similar to Lemma 1) leads to a contradiction. In the second case, by lifting properties we see that every cofibration must be surjective on connected components. It follows that every cofibrant object is empty, a contradiction.

Answer 2

The answer is no, there is no such model structure, at least not if the inclusion $i_0:\Delta^0\to \Delta^1$ is to be a cofibration.

EDIT: Tim Campion, in another answer, proves that there is no such model structure (without make any additional assumption).

Let's assume $i_0$ is a cofibration and reach a contradiction.

Since $i_0$ is also a weak equivalence (i.e. a simplicial homotopy equivalence in the sense of the question), it is a trivial cofibration: it has the left lifting property with respect to all fibrations.

Let $I(n)$ be the "subdivided interval", the nerve of the ordered set $0<\dots <n$. The inclusion $I(n-1)\to I(n)$ can be obtained by pushout from $i_0:\Delta^0\to \Delta^1$ and therefore it is also a trivial cofibration. Therefore for all $n\ge 1$ the inclusion of the left endpoint of $I(n)$ is a trivial cofibration, being the composition of (finitely many) trivial cofibrations. Call this map $i(n)$ $$ i(n):\Delta^0\to I(n). $$

It follows that the disjoint union $$\bigsqcup_{n\ge 1}\Delta^0\to \bigsqcup_{n\ge 1} I(n)$$ of all of the maps $i(n)$ is a trivial cofibration.

But this map is not a weak equivalence, because in order to make a "simplicial homotopy" from the identity map of $I(n)$ to the constant map to the left endpoint it requires $n$ actual simplicial homotopies: no finite sequence of simplicial homotopies can accomplish this for all $n$ simultaneously.

Answer 3

EDIT: The answer below suggests that there is a model structure on $sSet$ whose weak equivalences are the simplicial homotopy equivalences, but seems to have a problem. I'm going to leave what I wrote below, because it helps answer another interesting question, which is "What is the Barthel-Riehl $h$-model structure on $sSet$?" I now suspect that the $h$-model structure their machine outputs is actually the usual $q$-model structure. My reasoning is that the topological mapping space you get between two simplicial sets $X$ and $Y$ is $|Hom(X,Y)|$, i.e., the geometric realization of the internal hom in simplicial sets, defined by $Hom(X,Y)_n:= sSet(\Delta^n \times X,Y)$. So, when you trace through the meaning of a homotopy defined via these topological mapping spaces, I suspect you'll hit the same issue I pointed out in the last paragraph below, and discover that your weak equivalences are $\pi_*$-isomorphisms rather than simplicial homotopy equivalences.

A comment asked about the topological structure on $sSet$. What I had in mind follows. Let $Sing(-)$ denote the singular functor $Sing(I)_n:= \{\sigma: \Delta^n\to I\}$. The topological mapping spaces in $sSet$ are defined above. Given $T\in Top$ and $X,Y\in sSet$, define $T\otimes X:= Sing(T) \times X$. Then observe that $$|Hom(Sing(T)\times X,Y)| \cong |Hom(Sing(T),Hom(X,Y))| \simeq Top(|Sing(T)|,|Hom(X,Y)|)\simeq Top(T,|Hom(X,Y)|)$$ using this observation. The cotensoring is $X^T:=[Sing(T),X]$ and $$|Hom(X,Y^T)|\cong |Hom(X\otimes T,Y)| \cong |Hom(Sing(T)\times X,Y)|\cong |Hom(Sing(T),Hom(X,Y))|\simeq Top(|Sing(T)|,|Hom(X,Y)|) \simeq Top(T,|Hom(X,Y)|)$$ I think this is what Rognes meant by his Remark 3.9 that I cited below, and it was this structure I had in mind to use to define the notion of homotopy and homotopy equivalence for the $h$-model structure, following Barthel-Riehl. One of the problems is that $Sing(I)$ is much more complicated than $\Delta^1$.

Original answer:

The Strøm model structure is also known as the Hurewicz model structure, or $h$-model structure. Tobias Barthel and Emily Riehl gave general criteria for the existence of such model structures, in On the construction of functorial factorizations for model categories (published in AGT). In particular, Corollary 5.23 proves that any topologically bicomplete category $\mathcal C$ satisfying the monomorphism hypothesis admits an $h$-model structure, and Remark 5.20 points out that any locally presentable topologically bicomplete category satisfies the monomorphism hypothesis. Here $Top$ denotes the category of compactly generated weak Hausdorff spaces.

Observe that the geometric realization and singularization functors endow $sSet$ with the structure of a topological model category, and it is of course locally presentable. If you are in doubt, a reference is Remark 3.9 of these lecture notes by Rognes or Prop 5.6.15 of the book Equivariant stable homotopy theory and the Kervaire invariant problem, but dualizing the proof (using the pushout product formulation of the SM7 axiom). Hence, the category $sSet$ has an $h$-model structure. Its weak equivalences are the homotopy equivalences, as defined in Definition 5.1 of the Barthel-Riehl paper, using the tensor and cotensor with $Top$. It is easy to see that this is equivalent to the definition of a simplicial homotopy in your question.

Lastly, it is interesting to note that if you left-induce from the $h$-model structure on $Top$ along the geometric realization functor, you get the Quillen model structure on $sSet$ because a map $f: |X|\to |Y|$ is a homotopy equivalence if and only if it's a weak homotopy equivalence. See A.0.1 of the paper A necessary and sufficient condition for induced model structures. So, that approach does not produce a model structure whose weak equivalences are the simplicial homotopy equivalences.