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My question is very simple, but I wasn't able to find an answer in various sources. Cell-complexes are commonly presented using an inductive construction where $n$-cells are attached to $(n-1)$-cells, starting with the data of a collection $X^0$ of $0$-cells.

But sometimes, one will define a $2$-cell for instance, as being a sub-cell complex of something, raising the question: can a cell complex $0$-skeleton be empty? And more generally, can a cell-complex have empty skeletons until a dimension $k$?

It doesn't seem absurd to me that the answer should be yes, but it worries me to never see the case taken into account in the inductive construction.

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  • $\begingroup$ The empty space is a CW-complex with no $0$-cells. $\endgroup$ – Lord Shark the Unknown Jan 2 at 7:41
  • $\begingroup$ What do you mean by "cell complex"? Are you using that as a synonym for CW-complex? $\endgroup$ – Eric Wofsey Jan 2 at 7:44
  • $\begingroup$ No, I mean a cellular complex, which is basically the same as a CW complex with no closure-finiteness or weak-topology condition (see link ) $\endgroup$ – TryingToGetOut Jan 2 at 7:49
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Suppose $X$ is a nonempty cell complex and let $n$ be minimal such that $X$ has an $n$-cell. If $n>0$, then this $n$-cell has an attaching map $S^{n-1}\to X^{n-1}$ where $X^{n-1}$ is the $(n-1)$-skeleton of $X$. But by minimality of $n$, $X^{n-1}=\emptyset$. Since $S^{n-1}$ is nonempty, there are no maps $S^{n-1}\to\emptyset$, so this is a contradiction.

So, if $X$ is any nonempty CW-complex, it must have a $0$-cell. (Of course, the empty space is a CW-complex with no cells at all!)

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  • $\begingroup$ thanks, this seems to answer my question accurately! But is it necessary in the definition that the n-cells have attaching maps to the (n-1)-skeleton? Because the 0-cell will not have that for instance. $\endgroup$ – TryingToGetOut Jan 2 at 7:58
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    $\begingroup$ In this game, the boundary of the $0$-disc is empty so it attaches perfectly well to the empty $-1$-skeleton :-) @TryingToGetOut $\endgroup$ – Lord Shark the Unknown Jan 2 at 8:01
  • $\begingroup$ Yes of course.. Thanks! $\endgroup$ – TryingToGetOut Jan 2 at 8:07

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