## The Cech expansion: nerves of open covers

The Whitehead theorem in homotopy theory basically says that to fully understand the homotopy type of a CW-complex one only needs to know about the homotopy groups (really, the weak homotopy type). It is very easy to produce spaces (for instance, subsets of $R^2$ or $R^3$ to which Whitehead’s theorem doesn’t apply.

Heavier machinery is required to study the structure of these spaces which, locally, are more complicated than CW-complexes. One of the most traditional approaches to this problem is shape theory. A common technique in mathematics is to approximate complicated objects by simpler ones. For instance, approximating functions in calculus by Taylor polynomials of increasing degree. This is basically the approach of shape theory: approximate complicated spaces by simpler ones, in particular polyhedra (spaces built out of lines, triangles, tetrahedra, etc…). Borsuk, the inventor of shape theory, first used ANR’s to study the topology of compact metric spaces. A more modern approach pioneered by Segal and Mardesic [1] is categorical in nature and makes use of inverse systems of polyhedra.

Thought shape theory helps a great deal in our understanding of complicated spaces, it has its limits (pun intended); later on, we’ll run into some spaces where shape theory breaks down.

So where do we start?

Let’s look at the Cech expansion of a space $X$. This is supposed to let us approximate spaces by simpler ones. We’ll start with an open cover $\mathscr{U}$ of our space $X$. Even when you forget about the points in each set $U\in\mathscr{U}$, the open cover still gives a vague picture of $X$. For instance, take the following cover of the unit circle $S^1$ as a subspace of the plane.

Now forget about the points in the space.

The shape left still closely resembles that of a circle. We can even recover the circle from this data: replace each open set with a point. If two open sets intersect, draw a line segment between the two corresponding points.

We get a polyhedron homeomorphic to the circle. It isn’t usually true that we will get back the original space; this only happens in very special cases.

### The nerve of an open cover

Definition: An abstract simplicial complex is a set $S$ and a set $K$ consisting of finite subsets of $S$ such that if $A\in K$ and $B\subset A$, then $B\in K$. A vertex or 0-simplex is a singleton $\{s\}\in K$ and an n-simplex is a set $\{s_1,...,s_{n+1}\}\in K$ containing $n+1$ elements. The n-skeleton of $K$ is the set $K_n=\{A\in K||A|=n+1\}$

Now if $\mathscr{U}$ is an open cover of $X$, we construct an abstract simplicial complex $N(\mathscr{U})$ called the nerve of $\mathscr{U}$. An element of $\mathscr{U}$ is a finite set $\{U_1,U_2,...,U_n\}\subset \mathscr{U}$ such that $\bigcap_{i=1}^{n}U_i\neq \emptyset$. The geometric realization $|N(\mathscr{U})|$ is a geometric complex obtained by pasting simplices together using $N(\mathscr{U})$ as instructions.

Definition: The geometric realization of an abstract simplicial complex $K$ with vertex set $K_0$ is the topological space $|K|$ defined as a subset of the product $P=[0,1]^{K_0}$ of functions $f:K_0\to [0,1]$. In particular, $|K|$ is the set of functions $f\in P$ such that

1. $\{v\in K_0|f(v)>0\}\in K$  (in particular all but finitely many $f(v)$ are zero)

2. $\sum_{v} f(v)=1$

Give  $P$ the weak (or induced) topology so that $U$ is open in $P$ iff $U\cap [0,1]^{F}$ is open in $[0,1]^{F}$ for all finite sets $F\subset K_0$. $|K|$ is topologized with the subspace topology of $P$.

Sometimes we’ll write the simplex in $|K|$ spanned by vertices $s_1,...,s_n$ in $|K|$ as $[s_1,...,s_n]=\{f\in |K||f(s_i)>0\text{ for some }1\leq i\leq n\}$.

Back to the nerve:

While $|N(\mathscr{U})|$ is defined as the geometric realization of the nerve, it is a bit more intuitive to think of it in the following way.

Here is the cover:

An open cover

0 skeleton – A vertex of $N(\mathscr{U})$ is a set $U\in \mathscr{U}$

0-skeleton

1 skeleton – If $U_1\cap U_2 \neq \emptyset$, then place an edge (1-simplex) between $U_1$ and $U_2$.

1-skeleton

2 skeleton – If $U_1\cap U_2\cap U_3\neq \emptyset$, then there are three edges joining each pair of the vertices $U_1,U_2,U_3$. Place a triangle (or 2-simplex) so that the edges of the triangle match up with these three edges. In the picture, fill in the each empty triangle with a triangle.

3 skeleton – If $U_1\cap U_2\cap U_3\cap U_4\neq \emptyset$, attach a tetrahedron to fill in the boundary that exists from the four triangles.

tetrahedron

Input the tetrahedron into the place where it obviously goes (on the right).

In our example, we would stop here and leave it embedded in $R^3$ but, in general, you would continue to add higher dimensional simplices and give the resulting geometric simplicial complex the weak topology. In addition, the space might not be compact and open covers would typically contain infinitely many sets.

### Refinements:

The nerve $|N(\mathscr{U})|$ is supposed to be an “approximation” of the original space $X$. What if it is a bad approximation? Well…take a “closer look” at $X$ by covering $X$ with smaller open sets.

Definition: An open cover $\mathscr{V}$ is a refinement of another cover $\mathscr{U}$ if for each $V\in \mathscr{V}$ there is a $U\in \mathscr{U}$ such that $V\subseteq U$.

If $\mathscr{V}$ refines $\mathscr{U}$, then $|N(\mathscr{V})|$ is “larger” than $|N(\mathscr{U})|$ since there are more sets in $\mathscr{V}$. It makes sense to think of $|N(\mathscr{V})|$ as being a better approximation to $X$ since if we collapse the appropriate simplices of $|N(\mathscr{V})|$, we get back something homotopy equivalent to $|N(\mathscr{U})|$. This is captured in the next proposition.

Proposition: If  $\mathscr{V}$ is a refinement of another cover $\mathscr{U}$, there is a there is a simplicial map $p_{\mathscr{V}\mathscr{U}}:N(\mathscr{V})\to N(\mathscr{U})$. The map $|p_{\mathscr{V}\mathscr{U}}|:|N(\mathscr{V})|\to |N(\mathscr{U})|$ induced on geometric realizations is unique up to homotopy.

proof. First define $p_{\mathscr{V}\mathscr{U}}$ on vertices (i.e. elements of $\mathscr{V}$). If $V\in \mathscr{V}$ and $V\subseteq U_{V}$ for $U_{V}\in\mathscr{U}$, define $p_{\mathscr{V}\mathscr{U}}(V)=U_{V}$. If $V\cap V'\neq \emptyset$, then clearly $U_{V}\cap U_{V'}\neq \emptyset$ so we define $p_{\mathscr{V}\mathscr{U}}$ on the 1-simplex $[V,V']$ spanned by $V$ and $V'$ to the 1-simplex $\left[U_{V},U_{V'}\right]$ spanned by $U_{V}$ and $U_{V'}$. Any map defined in this way is called a projection.

The same goes for higher simplices; if $\bigcap_{i=1}^{n}V_i\neq \emptyset$, then $\bigcap_{i=1}^{n}U_{V_{i}}\neq \emptyset$ and we send the simplex $[V_1,...,V_n]$ to $[U_{V_{1}},...,U_{V_{n}}]$. This gives a well-defined simplicial map on the nerves.

Though it seems like there is a lot of freedom in defining a projection, it is not too hard to show that any two projections induce contiguous maps on geometric realizations. But contiguous maps of simplicial complexes are homotopic, proving the proposition.$\Box$

Note that if $\mathscr{W}$ is a refinement of $\mathscr{V}$ and $\mathscr{V}$ is a refinement of $\mathscr{U}$, the composition $p_{\mathscr{V}\mathscr{U}} \circ p_{\mathscr{W}\mathscr{V}}$ is a canonical map. Thus if $p_{\mathscr{W}\mathscr{U}}$ is another projection, then there is a homotopy $\left|p_{\mathscr{W}\mathscr{U}}\right| \simeq |p_{\mathscr{V}\mathscr{U}}| \circ |p_{\mathscr{W}\mathscr{V}}|$. Now if we let $[p_{\mathscr{V}\mathscr{U}}]$ denote the homotopy class of $|p_{\mathscr{V}\mathscr{U}}|$, we have strict equality $[p_{\mathscr{W}\mathscr{U}}] = [p_{\mathscr{V}\mathscr{U}}] \circ [p_{\mathscr{W}\mathscr{V}}]$. Therefore, since open covers of $X$ form a directed set $\mathcal{O}(X)$, we have an inverse system $\left(|N(\mathscr{U})|,[p_{\mathscr{V}\mathscr{U}}],\mathcal{O}(X)\right)$ of homotopy classes of nerves of covers.

Definition: For a paracompact Hausdorff space $X$, the Cech expansion of $X$ is the inverse system $\left(|N(\mathscr{U})|,[p_{\mathscr{V}\mathscr{U}}],\mathcal{O}(X)\right)$ in the homotopy category of polyhedra.

Of course, even if $X$ is not paracompact Hausdorff, you still get an inverse system; the problem with non-paracompact spaces is that it is much harder to relate $X$ to the inverse system without “enough” partitions of unity…hopefully more on this later.

### References:

[1] S. Mardsic and J. Segal, Shape theory, North-Holland Publishing Company, 1982.