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• CommentRowNumber1.
• CommentAuthorMirco Richter
• CommentTimeNov 7th 2011
• (edited Nov 7th 2011)

Suppose we have a simplicial set X and a m-truncated Kan simplicial set Y. Then how is it possible to construct $Hom_{Simpl}(X,Y)$ as a subset $H \subset Hom_{Set}(X_m,Y_m)$?

Since Severa used this in his work on the n-jet functor (for X the nerve of the pair groupoid over an arbitrary set), it should be possible. Nevertheless I can’t find an explicit construction including a proof that what he constructed is indeed $Hom_{Simpl}(X,Y)$.

By an explicit construction I mean something like: Let $f \in H$ be given, then the appropriate simplicial morphism $F$ is given by $F[n]= X_n \rightarrow Y_n$ as follows : ??? where the commutation with the face and degeneracy maps is seen as follows ??? … On the other side we that any simplicial morphism is given that way, because ???

….

So if someone could give me a proof (I think it will be an induction on something like $Hom_{Simpl}(Sk^n X,Y) \subset Hom_{Set}(X_n,Y_n)$ or $Hom_{Simpl}(Horn_j^n X,Y) \subset Hom_{Set}(X_n,Y_n)$ ) it would be great.

Likely this doesn’t work for arbitrary simplicial sets X, so another topic is to find the appropriate conditions on X .

Moreover this should be put into the nLab, too…

If nobody knows a proof it would be nice, if we could work it out together. At the end I will take the time to put in the nLab. Unfortunately my skills on simplicial sets are not good enough, to do it by myself.

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeNov 7th 2011
• (edited Nov 7th 2011)

Have a look at simplicial skeleton. If $Y$ is $m+1$-coskeletal then $Y \simeq \mathbf{cosk}_{m+1} Y$ and $\mathbf{cosk}$ (boldface) is the composition of chopping off the higher cells by a truncation map and then re-embedding by a right adjoint $cosk$ (non-boldface). Being a right adjoint, you can throw it over to $X$ and find that $Hom(X,Y)$ is the same as the hom $Hom(tr X , tr Y)$ between the truncations. Now it only remains to observe that every $k \leq m$-simplex can be regarded as a degenerate $m$-simplex (in several ways, but at least in one way). Use the simplicial identities to deduce that known the map on $m$-simplices defines it also on all lower dimensional simplices.

• CommentRowNumber3.
• CommentAuthorzskoda
• CommentTimeNov 7th 2011

I think this discussion should be classified into section Mathematics, Physics and Philosophy of $n$Forum – general discussions on $n$Lab where it is classified now is about general policies and issues about $n$Lab, if I understood the scheme correctly.

• CommentRowNumber4.
• CommentAuthorTobyBartels
• CommentTimeNov 7th 2011

Yes, I have moved it.

• CommentRowNumber5.
• CommentAuthorMirco Richter
• CommentTimeNov 8th 2011

I observe a lot of trouble with the expression of the Latex-Like expressions in my browser (Using Chromium on a Linux OS) All of them are typesetted wrong in a way. Mostly because they are written ’to high’ relative to the text-line. Moreover in the last comment of Urs there are no expressions at all. I just see a repeated abstract icon ..

If you want I could send you a screen shot.

Unfortunately this way your comment is unreadable to me.

• CommentRowNumber6.
• CommentAuthorMirco Richter
• CommentTimeNov 8th 2011
• (edited Nov 8th 2011)

So what Urs said just describes how to construct the simplicial map from $Hom(X_m,Y_m)$ but of course not every map $f \in Hom(X_m,Y_m)$ extends to a simplicial morphism. In particular it must map degenerated simplices to degenerated simplices “of the same level of degeneration” and maps common faces to common faces.

Ok but that’s just tells us that there is a way.

So HOW TO construct this subset? Because this is a finite recursion there must be some kind of algorithmic process starting at $Hom(X_0,Y_0)$ and terminating at $Hom(X_m,Y_m)$

(Maybe this is just a philosophical disput on what a solution to such a problem means. Just to say that there is a solution is in my mind just a prestep to giving a solution, at least if a solution is constructable)

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeNov 8th 2011
• (edited Nov 8th 2011)

So HOW TO construct this subset?

First consider the adjunction isomorphism

$Hom_{sSet}(X,Y) \simeq Hom_{sSet}(X, cosk tr Y) \simeq Hom_{sSet_{n+1}}(tr X, tr Y)$

and then the defining inclusion $Hom_{sSet_{n+1}}(tr X, tr Y) \hookrightarrow Hom_{Set}(X_{n+1}, Y_{n+1})$.

Is that not what you are after?

• CommentRowNumber8.
• CommentAuthorMirco Richter
• CommentTimeNov 8th 2011
• (edited Nov 8th 2011)

Finally yes… But the question is more than that. Suppose I have $Hom_{sSet}(sk^0X,Y)$ then because $sk^0X$ has only degenerated simplices in dimension $\geq 1$ we have $Hom_{sSet}(sk^0X,Y) \simeq Hom_{Set}(X_0,Y_0)$

Ok. That’s the beginning of the induction.

Now suppose we have $Hom_{sSet}(sk^k X,Y) \hookrightarrow Hom_{Set}(X_k,Y_k)$ is it possible and if yes HOW, to get $Hom_{sSet}(sk^{k+1} X,Y) \hookrightarrow Hom_{Set}(X_{k+1},Y_{k+1})$ from this data?

Then this would be the induction step.

After doing this $m$ times we could stop since we have what Urs wrote in his last post.

Maybe it is a bit imprecise to call this constructive, but because I assume we don’t know $Hom_{sSet}(X,Y)$ in the first place, it is surely more easy (from a technical point of view) to describe $Hom(X_0,Y_0)$ first and then recursively construct the full-fledged set $Hom_{sSet}(X,Y)$ from this data in a finite recursive process. Especially if there is additional structure on it, like being a manifold and one wants to construct charts or things like that.

Maybe I can’t make myself clear here.

• CommentRowNumber9.
• CommentAuthorTim_Porter
• CommentTimeNov 9th 2011
• (edited Nov 9th 2011)

Is it not the case that there will be an obstruction theoretic argument needed to get from one level to the next? You have to be able to say where any non-degenerate k+1 simplices in X are to be sent and if their boundaries are sent to empty k-spheres in Y there is no extension of the construction so far. (Am I understanding your problem?) The inclusion of $sk^k X$ into $sk^{k+1}X$ is a cofibration so there is an induced fibration on the homs into Y. You need to be able to analyse that fibration.

• CommentRowNumber10.
• CommentAuthorMirco Richter
• CommentTimeNov 10th 2011
• (edited Nov 10th 2011)

Tim can you tell a little more about the induced fibration on the ’Homs”. Do you mean that something like $Hom(sk^*X,Y)$ with $Hom(sk^* X,Y)_n := Hom(sk^n X,Y)$ is a fbration?

• CommentRowNumber11.
• CommentAuthorTim_Porter
• CommentTimeNov 10th 2011

If I remember rightly if $A\to X$ is a cofibration, then for any $Y$, $Y^X\to Y^A$ is a fibration (provided $Y$ is fibrant i.e. Kan). (I have not the source in front of me, so may be forgetting some conditions or, as I have not had that much coffee this morning I may be misremembering! So do check.) Now apply it to the inclusion of the k-skel into the k+1-skel. (I find that SLN 100 by Artin and Mazur has a good short summary of this sort of thing and the relation with coskels as well. Again I find that the basic inductive ‘up the skeleton’ constructions have greater intuitive use when in a geometric context, even if after understanding that approach a neat adjointness argument is the one I eventually write down and type up! In this case I think they have great value for giving additional depth to the categorical arguents.)

• CommentRowNumber12.
• CommentAuthorUrs
• CommentTimeNov 10th 2011

If I remember rightly

Yes. I have added more dicussion of this point to simplicial model category in the section Properties – Enrichment, tensoring and cotensoring