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Let $A, X$ be simplicial sets, and suppose there's an inclusion $A \longrightarrow X$. Geometrically realizing the inclusion map, we get a pair of spaces $(\mathcal{A}, \mathcal{X})$. I want to find a simplicial set whose geometric realization has the homotopy type of $\mathcal{X} \setminus \mathcal{A}$.

Here are my thoughts so far:

  1. If we try to remove from $X$ all simplices that are also in $A$, we run into immediate trouble: other simplices in $X$ probably have simplices of $A$ as their faces. Somehow we need to remove the simplices of $A$ and then adjoin new boundary simplices ``as freely as possible.''
  2. Ideally such a construction would be functorial. There's no way that the domain can be the category of pairs, however, since there won't be induced maps after subtracting. Instead, it seems like the source category should be more like a twisted arrow category (although this doesn't seem to work either).

The case I'd really like to get working is where $X$ is a $G$-simplicial set with finitely-many non-degenerate simplices in each degree and $A$ is the points which have nontrivial stabilizer, but I'm interested in the general case as well.

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  • $\begingroup$ John, what precisely do you mean by X\A and taking its `geometric realization'. It seems you mean the simplices of X not in A, which as you observe is not a simplicial set, so its geometric realization has no obvious meaning. You are desperately trying to avoid X/A. In your equivariant example, presumably G is a group acting simplicially and A is the subcomplex of simplices with nontrivial stabilizer. Then you might replace X\A by the maximal G-free subcomplex of X, which maybe makes sense. $\endgroup$
    – Peter May
    Commented Apr 15, 2013 at 1:27
  • $\begingroup$ Dear Professor May, I meant geometrically realizing A and X first and then performing subtraction in the category of spaces. $\endgroup$
    – John Wiltshire-Gordon
    Commented Apr 15, 2013 at 1:31
  • $\begingroup$ The singular simplicial set of the space $|X| \setminus |A|$ has the property you ask for, namely its geometric realization is homotopy equivalent to $|X| \setminus |A|$. Perhaps you want some additional constraints? $\endgroup$
    – Chris Schommer-Pries
    Commented Apr 15, 2013 at 14:03
  • $\begingroup$ @Chris Schommer-Pries: Good point, but I'd like a combinatorial answer. In particular, if there are finitely many simplices before, I'd like finitely many after. $\endgroup$
    – John Wiltshire-Gordon
    Commented Apr 15, 2013 at 15:42

2 Answers 2

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Here are some perhaps silly ideas.

(i) Think of simplicial sets as presheaves on $\Delta$, now look up when a subpresheaf can be subtracted. (E.g. is there a universal property involved?) What os the subobject classified in the presheaf topos here? I am sure that these things are known, but possibly do not have a simple answer.

(ii) One related point from Peter May's answer is that $X\setminus A$ should be the maximal subsimplicial set of $X$ with trivial intersection with $A$. The condition on the realisations is then a possible red herring.

(iii) A final point is that if you are handling a simplicial complex (say a PL manifold, with a given triangulation) there would be questions of subdivision, and if I remember rightly some sort of regular neighbourhoods, and I think that this is geometrically more significant that merely looking at the complement of $|A|$ in $|X|$.

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The 0-simplices of Sd(X) are the simplices of X. Then X\A gives a subset of Sd_0(X), how about the subspace of Sd_.(X) spanned by those?

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    $\begingroup$ I do not know if this works or not. $\endgroup$
    – John Wiltshire-Gordon
    Commented Apr 20, 2013 at 22:28

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