Proving Invariance of CW Complex Dimension

Question

John Lee's Introduction to Topological Manifolds defines the dimension of a CW complex $X$ as the largest dimension of a cell in $X$. The author claims "The fact that this well defined depends on the theorem of invariance of dimension."

I'm assuming he is referring to invariance of dimension for manifolds, since it's the only invariance of dimension result mentioned up to that point, but since not all CW complexes are manifolds, I don't immediately see how the proof would go. Is it meant to be easy, or does it take some more machinery?

Answer 1

I think John Lee meant to use the full power of invariance of domain. I don't think that it can be deduced simply from the fact that homeomorphic manifolds have equal dimensions (unless "invariance of dimension" means something else). I believe a stronger variant is needed, in particular that there is no continuous injective map $\mathbb{R}^n\to\mathbb{R}^m$ if $n>m$.

More generally, assume that $f:X\to Y$ is a homeomorphism between CW complexes, with fixed CW structures. Assume that $X$ has an $n$-cell. This means that we have an embedding of a closed ball $i:D^n\to X$. $D^n$ is compact and thus so is the image of $f\circ i$. In particular it has to be contained in a finite subcomplex $Y'$. Take an open cell $e$ of maximal dimension $m$ in $Y'$ which intersects the image of $i$ (such cell has to exist). First of all $e$ is homeomorphic to $\mathbb{R}^m$, as all open cells are. Secondly this cell is open in $Y'$, as all maximal open cells are. Thus $(f\circ i)^{-1}(e)$ is open in $D^n$, in particular it intersects the interior of $D^n$. And so we can restrict $i$ to an open ball, which is homeomorphic to $\mathbb{R}^n$. Combining everything together we obtain a continuous injective map $\mathbb{R}^n\to\mathbb{R}^m$, which by the invariance of domain can only happen when $n\leq m$.

Concluding: if $X$ contains a cell of dimension $n$, then $Y$ contains a cell of dimension at least $n$. For that we only need $f$ to be injective. The vice versa also holds when $f$ is a homeomorphism, because we can apply the previous result to $f^{-1}$, more precisely: if $Y$ contains a cell of dimension $m$ then $X$ contains a cell of dimension at least $m$. Both of course imply that they have equal dimension (finite or not), and so the dimension is a topological invariant.

Answer 2

User @freakish is right that the most straightforward way to prove this is with invariance of domain. But with a little more work, it can actually be done using only invariance of dimension (the theorem that a nonempty topological space cannot be both an $m$-manifold and an $n$-manifold for $n\ne m$). Here's a proof.

Let $X$ be a topological space with a CW decomposition. I claim first that if $U$ is a precompact open subset of $X$ homeomorphic to an open subset of $\mathbb R^n$, then $\overline U$ is contained in the $n$-skeleton of $X$. Because $\overline U$ is compact, it is contained in a finite subcomplex of $X$. Let $X'$ be such a subcomplex of minimal dimension $m$. At least one of the open $m$-cells of $X'$ must intersect $\overline U$ -- if not, we can just discard all the open $m$-cells and obtain a finite subcomplex of lower dimension that contains $\overline U$. Let $e'$ be such an $m$-cell. As a top-dimensional cell of $X'$, $e'$ is open in $X'$. On the other hand, since $U$ is open in $X$ and contained in $X'$, $U$ is open in $X'$. The intersection $\overline U\cap e'$ is nonempty and open in $\overline U$, so it must contain points of $U$. Thus $U\cap e'$ is nonempty, open in $U$ (hence homeomorphic to an open subset of $\mathbb R^n$), and open in $e'$ (hence homeomorphic to an open subset of $\mathbb R^m$). Therefore, by invariance of dimension, $m=n$, and thus $\overline U$ is contained in the $n$-skeleton.

Now suppose $X$ has two CW decompositions, $\scr E$ and $\scr E'$, with $\scr E$ of dimension $n$. I'll show that all of $X$ is contained in the $n$-skeleton of $\scr E'$, which implies that every other CW decomposition is finite-dimensional and of dimension at most $n$; and reversing the roles of $\scr E$ and $\scr E'$ shows the dimension is exactly $n$.

First, every open $n$-cell of $\scr E$ is open and precompact in $X$, and thus its closure is contained in the $n$-skeleton of $\scr E'$ by the argument above. Let $V$ be the union of the closures of all the $n$-cells of $\scr E$. Since the $n$-skeleton of $\scr E'$ is closed in $X$, it follows that the $n$-skeleton of $\scr E'$ contains $\overline V$.

All that remains is to show that the open cells of $\scr E$ that are not contained in $\overline V$ are also contained in the $n$-skeleton of $\scr E'$. Let $e$ be such a cell of dimension $n-1$, and let $U = e\cap (X\smallsetminus \overline{V})$. Then $U$ is nonempty and open in $e$, hence open in $\overline e$. Its intersection with the closure of every other cell of $\scr E$ is empty, so it is also open in $X$. The argument above shows that $\overline U$ is contained in the $n$-skeleton of $\scr E'$. Apply this argument to show that the closures of all the $(n-1)$-cells is contained in the $n$-skeleton, as is the closure of their union; and then argue similarly for the $(n-2)$-cells, etc.