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 $n=\infty$.


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 $m\geq n$.

  • $\begingroup$ 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

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