5
$\begingroup$

This is from page 204 of Rotman's An Introduction to Algebraic Topology. After some elementary definitions and facts about CW copmlexes, exercise 8.27 asks:

Define the dimension of a CW complex $(X,E)$ to be $$\mbox{dim }X=\mbox{sup}\{\mbox{dim}(e):e\in E\}.$$ If $E'$ is another CW decomposition of $X$, show that $(X,E)$ and $(X,E')$ have the same dimension.

My attempt: To distinguish, let us write $\mbox{dim }(X)$ for $(X,E)$ and $\mbox{dim' }(X)$ for $(X,E')$. First consider the case when both dimensions are finite, say $\mbox{dim }(X)=m$ and $\mbox{dim' }(X)=n$ and assume for contradiction that $m<n<\infty$ holds. Choose an $n$-cell $e'$ in $(X,E')$ which must be open in $X$ since it is of largest dimension in $E'$. Viewing $X$ as $(X,E)$, $e'$ meets some of the cells in $E$ nontrivially, so let $e$ be such a cell in $E$ with maximal dimension. We now have $k=\mbox{dim}(e)\leq m<n$.

Now what I want to show is that the intersection $e'\cap e$ must be open in $X$, but I can't proceed any further. I think that after this fact is established, one can use the Invariance of Domain to conclude easily that $k<n$ gives a contradiction: $e'\cap e$ is homeomorphic to some open subset of $\mathbb{R}^{n}$ and to some open subset of $\mathbb{R}^{k}$ at the same time.

Is it true that for my choice of $e',e$, the intersection $e'\cap e$ is open in $X$? (I strongly believe that this is the case!) And how should one take care of the case when one of the dimension is infinite, say $n=\infty$? (for in that case we can't choose open $e'$ so easily...). Any help would be appreciated.

Please enlighten me.

EDIT: OK, I figured out that the intersection must be open in $X$. So the only remaining question is about how to handle the case $n=\infty$.

$\endgroup$
2
$\begingroup$

Here is one way to give a manifestly homeomorphism-invariant definition of dimension for CW-complexes, and thus show that the definition is independent of the chosen CW structure. The idea is basically the same as your approach, but I find it a bit cleaner. I claim that $X$ has dimension $\geq n$ iff there exists an embedding from an $n$-simplex $\Delta^n$ to $X$.

Clearly if $X$ has a cell of dimension $\geq n$, then we can embed $\Delta^n$ in the interior of that cell and thus in $X$. Conversely, suppose $i:\Delta^n\to X$ is an embedding. Since $\Delta^n$ is compact, the image of $i$ is contained in a finite subcomplex of $X$. Let $e\subseteq X$ be an open cell of maximal dimension (say, $m$) which intersects the image of $i$. By maximality of the dimension of $e$, $i^{-1}(e)$ is open in $\Delta^n$. In particular, it contains an open ball in the interior of $\Delta^n$. Restricting $i$ to this open ball and identifying $e$ with $\mathbb{R}^m$, we get an embedding of $\mathbb{R}^n$ into $\mathbb{R}^m$. By invariance of domain, this is only possible if $m\geq n$.

$\endgroup$
  • $\begingroup$ Could you explain the part "Let $e\subseteq X$ be an open cell of maximal dimension (say, $m$) which intersects the image of $i$"? Do you mean that choosing a cell $e$ with maximal dimension intersecting the image must be open in $X$? It would be nice if you could explain where the maximality was used... $\endgroup$ – Dilemian Aug 27 '16 at 3:53
  • $\begingroup$ By "open cell" I just mean I'm only taking the interior of the cell, not the entire image of the map $\Delta^m\to X$ (this appears to be what you just call as "cell"). Maximality of $m$ implies that $e$ is open as a subset of the subspace of $X$ consisting of the cells that intersect the image of $i$. $\endgroup$ – Eric Wofsey Aug 27 '16 at 4:04
  • $\begingroup$ I'm sorry, but why does the maximality imply openness? $\endgroup$ – Dilemian Aug 27 '16 at 4:32
  • $\begingroup$ For instance, let $Y\subseteq X$ be the smallest subcomplex of $X$ containing the image of $i$ (so it contains the cells that intersect the image of $i$, as well as the cells that the closures of those cells intersect, and so on). Then $e$ is a maximal-dimensional cell of $Y$, and hence an open subset of $Y$. $\endgroup$ – Eric Wofsey Aug 27 '16 at 4:53
  • $\begingroup$ Thank you for your kind explanation. I think I understand the argument now. $\endgroup$ – Dilemian Aug 27 '16 at 7:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.