# Characteristic map of a n-cell in a CW complex

I have a problem in understanding the purpose of the definition of a CW complex. What really would help me is to understand the following: Let $\sigma$ be a n-cell and $\Phi_\sigma:\mathbb D^n \to X$ be the characteristic map. Is there any relation between $\Phi_\sigma(\mathbb S^{n-1})$ and $\partial \sigma$ following directly from the Definition? I know, for example, that $\Phi_\sigma(\mathbb S^{n-1})\subset X^{n-1}$, $X^{n-1}$ being the $(n-1)$-Skeleton, but what does this tell me about the boundary of $\sigma$?

If the cell $e^n_\alpha$ is defined as $\Phi_\alpha(\text{int}D^n)$, then you have the following relation:
Since the characteristic map $\Phi_\alpha$ is continuous, you always have $\Phi_\alpha(D^n)=\Phi_\alpha\left(\overline{\text{int}D^n}\right)\subseteq\overline{e^n_\alpha}$. On the other hand, once you have proven that CW-complexes are Hausdorff (and, more generally, normal), you know that compact subsets are closed. Therefore $\Phi(D^n)$ is closed, and since it contains $e^n_\alpha$, it also contains the closure $\overline{e^n_\alpha}$. So in the end, $\Phi_\alpha(D^n)=\overline{e^n_\alpha}$.

But beware that $\partial e_\alpha$ is in general not the same as $\Phi_\alpha\left(S^{n-1}\right)$. For example, $\partial e_\alpha$ can be the cell $e_\alpha$ itself if it's a $2$-cell with a $3$-cell glued to it.

• Of course, thx! So, at least $\overline{e^n_\alpha}\e^n_\alpha$ is contained in $X^{n-1}$, that's very encouraging :) – user83496 Aug 26 '13 at 15:58
• Yes, because $\overline{e^n_\alpha}-e^n_\alpha$ is the same as $\Phi_\alpha(S^{n-1})$ since $\Phi(\partial D^n)$ and $e^n_\alpha$ are disjoint. – Stefan Hamcke Aug 26 '13 at 16:15
• @StefanHamcke, I am reading Introduction to Homotopy Theory by Martin Arkowitz. Exercise 1.16 of the book asks the reader to prove that the boundary of $e_{\alpha}$ is equal to $\Phi(S^{n-1})$. Does it contradict the first sentence of the second paragraph of your answer? – Zuriel Oct 28 '14 at 12:48
• @Zuriel: Sorry for the late answer, I had to finish my diploma thesis. The boundary of a cell $e_\alpha$ must contain $\overline{e_α}-e_α=\Phi\left(S^{n-1}\right)$, but it could be all of $e_α$, for example when $e_α$ is the $1$-cell in a complex consisting of a $1$-cell whose end points are attached to a point, and a $2$-cell which is a disk attached along its boundary via a homeomorphism to the circle $\overline{e_α}$. – Stefan Hamcke Dec 4 '14 at 15:15

I think it helps to have some pictures. Here is a picture of a $$1$$-dimensional cell complex. For a $$2$$-dimensional example, try attaching a $$2$$-cell to the unit interval $$[0,1]$$, itself thought of as a cell complex with two $$0$$ cells and one $$1$$-cell. (The mouth in the face.) Here are two examples of attaching a $$2$$-cell. In the first the attaching map takes the boundary of the $$2$$-disc to the point $$1$$. In the second, it takes the boundary of the $$2$$-disc to somewhere in the middle of the unit interval.

These pictures are taken from Topology and Groupoids (colour in the e-version). I think it helps also to develop some of the general theory of attaching a space $$X$$ to a space $$B$$ by means of a map $$f:A \to B$$ on a closed subspace $$A$$ of $$X$$.