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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeApr 12th 2013

brief entry model structure on semi-simplicial sets, just in order to record a recent note by Benno van den Berg.

• CommentRowNumber2.
• CommentAuthorZhen Lin
• CommentTimeFeb 16th 2016

There’s also this post of Peter May’s that says that there is no possible model structure with weak equivalences defined by geometric realisation. Has anyone checked Benno’s assertion that his Quillen adjunction is a Quillen equivalence? That would be a quite interesting situation – then semisimplicial sets would model homotopy types in two different ways!

1. I just took a quick look, and I don’t think that $(i^{*}, i_{*})$ defines a Quillen equivalence.

Indeed, since $i_{*}$ preserves weak equivalences by definition (and in particular thus takes fibrant replacements to weak equivalences), we would have to have, by the 2-out-of-3 property, that the adjunction morphism $X \rightarrow i_{*}i^{*}(X)$ is a weak equivalence of simplicial sets for any simplicial set $X$.

Now $i_{*}(Y)$, for any semi-simplicial set $Y$, has the following explicit description, if I am not mistaken: the set of $n$-simplices is the set of morphisms from $i^{*}(\Delta^{n})$ to $Y$; and the faces and degeneracies come from pre-composing with $i^{*}$ applied to the usual face and degeneracy arrows.

Let us take $X$ to be the simplicial set which is $\Delta^{1}$ together with a (non-degenerate) loop glued on at one of the two $0$-simplices. Then $i_{*}i^{*}(X)$ is, as far as I see, a simplicial set with three 0-simplices and six non-degenerate 1-simplices, which form a triangle, where there are four edges joining one of the pairs of vertices. There are no non-degenerate $n$-simplices for any $n \geq 2$.

In particular, the geometric realisation of $i_{*}i^{*}(X)$ is certainly not homotopy equivalent to the geometric realisation of $X$, which is homotopy equivalent to a circle.

It is rather easy to overlook something when working out what $i_{*} i^{*}(X)$ is explicitly, so I would definitely recommend anyone interested to check for themselves that they agree that my calculation is correct.

• CommentRowNumber4.
• CommentAuthorDmitri Pavlov
• CommentTimeFeb 17th 2016
• (edited Feb 17th 2016)

Re #3: Is i^*X fibrant in your example? If not, it must be fibrantly replaced first, otherwise you don’t get the correct (i.e., derived) unit X→Ri_* i^*X (here X is cofibrant), which is the map that must be a weak equivalence.

2. This is what the second paragraph of #3 addresses.

• CommentRowNumber6.
• CommentAuthorZhen Lin
• CommentTimeFeb 17th 2016

I agree that $i_* Y$ has $n$-simplices given by morphisms $i^* \Delta^n \to Y$, but I don’t follow your example. Where do you get the third vertex? There are the two obvious ones, and I guess you might try to define a third one that maps the “degenerate” edge to the loop, but then what do you do with the “degenerate” 2-simplex?

3. On reflection, I believe that you are correct, and that there are only two vertices. I had indeed thought that one could send the single edge of $i^{*}(\Delta^{0})$ to the loop of $X$, but you are right that there is nowhere then to send the single 2-simplex of $i^{*}(\Delta^{0})$.

With this correction, I think that $i_{*}i^{*}(X)$ has exactly three non-degenerate 1-simplices. In addition to the degenerate cases, one can either:

i) send all three 1-simplices of $i^{*}(\Delta^{1})$ to the non-degenerate loop;

ii) send one to the degenerate loop, one to the non-loop, and one to the other degenerate loop;

iii) or send one to the non-degenerate loop, one to the non-loop, and one to the degenerate loop which does not have the same 0-simplex as the non-degenerate loop.

One can see that there are no non-degenerate 2-simplices by noting that a morphism $i^{*}(\Delta^{2}) \rightarrow i^{*}(X)$ would correspond to a morphism $i_{!}i^{*}(\Delta^{2}) \rightarrow X$, where $i_{!}$ is the left adjoint to $i^{*}$, and there is obviously no such morphism except one corresponding to a degeneracy.

Then the geometric realisation of $i_{*}i^{*}(X)$ is a figure of eight (a wedge sum of a pair of circles), which is not homotopy equivalent to a circle, as needed.

• CommentRowNumber8.
• CommentAuthorKarol Szumiło
• CommentTimeFeb 17th 2016
• (edited Feb 17th 2016)

I believe that the unit $X \to i_* i^* X$ is always a weak equivalence. We have $(i_* i^* X)_n = \mathsf{sSet}(i_! i^* \Delta[n], X)$ and the unit is induced by the counit $i_! i^* \Delta[n] \to \Delta[n]$ which is a simplicial homotopy equivalence as I explained here. A theorem of Latch, Thomason and Wilson (Thm 4.1 in Simplicial Sets From categories) then implies that the unit $X \to i_* i^* X$ is indeed a weak equivalence.

Since $i_*$ creates weak equivalences, it follows by triangular identities that counit $i^* i_* Y \to Y$ is also a weak equivalence for all $Y$. Thus it seems to me that this is indeed a Quillen equivalence.

The model structure is quite bizzare in any case. For example all finite semisimplicial sets are weakly equivalent to the empty one.

4. You may well be right. I would have to think some more about your argument, but in my description of the 1-simplices of $i_{*}i^{*}(X)$ in #7, I again overlooked a problem with where the 2-simplices of $i^{*}(\Delta^{1})$ are to be sent, I believe.

I would like to think about another example or two, but am beginning now to get some hands-on feeling for how we could have a Quillen equivalence.

The model structure is quite bizzare in any case. For example all finite simplicial sets are weakly equivalent to the empty one.

Yes, indeed!

• CommentRowNumber10.
• CommentAuthorRichard Williamson
• CommentTimeFeb 17th 2016
• (edited Feb 17th 2016)

Having thought about it a little more, I think the point is that the only way that $i_{*}i^{*}(X)$ can acquire new $n$-simplices as compared to $X$ is by virtue of $X$ containing something contractible, so that, up to homotopy equivalence, there is no difference. One might be able to make this formal by an argument involving homotopy colimits.

Indeed, in order to obtain a new 0-simplex, for instance, one has to at some point have an $n$-simplex whose faces are all degenerate, and then one has to have an $(n+1)$-simplex whose faces are all this $n$-simplex, and so on for all $m \geq n$.

I think the simplest example where $i_{*}i^{*}(X)$ is not simply isomorphic to $X$ is the simplicial set with a single non-degenerate $n$-simplex for every $n \geq 0$, whose faces are all the the non-degenerate $(n-1)$-simplex. To describe $i_{*}i^{*}(X)$ explicitly in this case would be rather difficult, but I think that, if one explores what it is a little, one can see how it could be homotopy equivalent to $X$.

• CommentRowNumber11.
• CommentAuthorKarol Szumiło
• CommentTimeFeb 17th 2016

I admit that I haven’t read your construction carefully so I don’t know where the mistake is, if there indeed is one. But I would guess that your suspicion form #9 is right. Semisimplicial sets $i^* \Delta[n]$ are not finitely generated so you cannot define a map out of one just by describing its behavior on a few simplices. If you want to do it carefully, things get rather tedious.

• CommentRowNumber12.
• CommentAuthorRichard Williamson
• CommentTimeFeb 20th 2016
• (edited Feb 20th 2016)

Yes, I am rather sure that there is a problem with the 2-simplices (and higher) with my proposed example. But I found the consideration of it, after the remarks of Zhen Lin and yourself, rather helpful in getting a hands-on feeling for why $i_{*}i^{*}(X)$ might be weakly equivalent to $X$. The remarks of #10 were intended to give some kind of summary of my conclusions, but probably one would need to try to construct a counter-example oneself, and see why it does not work, to appreciate them!

Samuel Mimram

• CommentRowNumber14.
• CommentAuthorUrs
• CommentTimeJun 26th 2021

• Simon Henry, Theorem 5.5.6 of: Weak model categories in classical and constructive mathematics, Theory and Applications of Categories, Vol. 35, 2020, No. 24, pp 875-958. (arXiv:1807.02650, tac:35-24)
• CommentRowNumber15.
• CommentAuthorUrs
• CommentTimeJun 26th 2021