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Let $K$ be an abstract simplicial complex on the (finite) vertex set $V$. The geometric realization $|K|$ is typically defined (see Spanier's book for instance) as the collection of functions $\alpha:V \to \mathbb{R}$ so that (a) the support of each $\alpha$ is a simplex, and (b) the sum $\sum_{v \in V}\alpha(v)$ equals $1$. Now each (closed) simplex $\sigma$ is realized as the collection of $\alpha \in K$ so that $\alpha(v) \neq 0$ implies $v \in \sigma$. From this one knows the star of each simplex.

A simplicial approximation of $f:|K| \to |L|$ is a simplicial map $g:K \to L$ so that $f(\text{star }\sigma) \subset \text{star }g(\sigma)$ for each simplex $\sigma \in K$. It is a standard result that the Piecewise Linear map induced by $g$ is homotopy equivalent to $f$

Now consider the case where $K$ is not abstract, but rather $V$ is an open cover of some topological space $X$. So, each simplex corresponds to an actual topological space, i.e., a non-empty intersection of some finite open sets in $X$. Let's call this $X_\sigma$.

My question is this:

What is the relation between $|K|$ and $X$, more specifically between $|\sigma|$ and $X_\sigma$ for each simplex $\sigma \in K$?

Here is some idea of what type of answer I am hoping for:

In the case where $X$ is paracompact and $V$ is a contractible cover, the nerve theorem applies and I know that $X$ and $|K|$ are homotopy equivalent. But is there a more general relationship between these two notions of realization of which the Nerve theorem is a consequence?

Furthermore, is there some functoriality to the nerve theorem? That is, assume you are given contractible covers $U$ and $V$ of $X$ and $Y$ generating the nerves $K$ and $L$. Given a function $f : X \to Y$ and a simplicial map $g:K \to L$, is there some magic analogoue of the star condition like $f(X_\sigma) \subset Y_{g(\sigma)}$ that makes $g$ induce a map homotopy equivalent to the composite $|K| \to X \to Y \to |L|$ where the maps on the edge come from the nerve theorem and the map in the middle is $f$?

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Regarding

What is the relation between $|K|$ and $X$, more specifically between $|\sigma|$ and $X_\sigma$ for each simplex $\sigma \in K$?

It seems to me that one can build an intermediate space $Y$ and a diagram $$ |K| \leftarrow Y \rightarrow X $$ which is natural in both the cover and in $X$ (where if we have a map $X\to X'$ the covering for $X$ should be the inverse image of the covering elements for $X'$). The space $Y$ is given by the realization of the nerve of the topological poset $\cal P$ whose elements are pairs $(U,x)$ in which $U$ is a finite collection of intersection of open sets in the covering $V$, and $x$ is a point of $U$.

There are then forgetful maps $Y \to |K|$ as well as $Y \to X$.

To see this, note that $X$ can be regarded as a topological category (or poset) whose objects are points of $X$ and only identity morphisms. Then ${\cal P} \to X$ is just the forgetful functor and it induces the map $Y \to X$ on realization (the realization of $X$ when considered as a topological poset is $X$ as a space). There is also a forgetful functor from ${\cal P}$ to the nerve of the covering which induces the map $Y \to |K|$.

If every non-empty finite intersection of members of $V$ is contractible, then the maps $Y \to |K|$, $Y \to X$ are weak equivalences.

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  • $\begingroup$ In your pairs $(U,x)$ do you require $x \in U$? $\endgroup$
    – Pinying
    Commented Nov 22, 2012 at 11:40
  • $\begingroup$ yes. I've edited to take that into account. $\endgroup$
    – John Klein
    Commented Nov 22, 2012 at 12:36
  • $\begingroup$ The map $Y\to X$ looks like it's going to be a weak equivalence in general, so that (in an up-to-homotopy way) you have an indirect map $X\to |K|$ in any case. $\endgroup$ Commented Nov 22, 2012 at 14:20
  • $\begingroup$ @Tom: That's a cofinality type argument, right? $\endgroup$
    – John Klein
    Commented Nov 23, 2012 at 4:50
  • $\begingroup$ Thank you. If you have any ideas about the functoriality question, please also write them down. In any case, I will accept this nice answer! $\endgroup$
    – Pinying
    Commented Nov 26, 2012 at 2:12

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