I am trying to find a proof that any finite cell complex $X = \bigcup_{i=0}^nX_n$ is homotopy equivalent to a regular complex $Y$, where a complex is regular if each attaching map is an embedding.

I cannot find a reference with a proof of this fact. I believe that I've seen it given as an exercise. It seems that $Y$ will need to have many more cells than $X$. But I'm having trouble defining the attaching maps for $Y$.

Let $\varphi_{\alpha} : S^k \to X_k$ be an attaching map for $X$. If I remember correctly, the exercise said to consider the map $\psi_{\alpha} : S^k \to X_k \times D^{k+1}$ given by $\psi(x) =(\varphi(x),x). $ Certainly this would be injective, but it doesn't make much sense as an attaching map for $Y$. So I'm not sure how to use it to construct $Y$.

Any help would be much appreciated.

NOTE: I am aware that Hatcher has a proof that every cell complex is homotopy equivalent to a simplicial complex and that any simplicial complex is regular. He relies on the simplicial approximation theorem and modifies constructions such as the mapping cone specifically to prove equivalence to a simplicial complex. I am asking for a direct proof where all we wish to prove is homotopy equivalence to a regular complex. I believe that one need not have to go through the machinery Hatcher does for my question.

  • $\begingroup$ I doubt that there is a really "elementary" proof. But of course I may may wrong. $\endgroup$
    – Paul Frost
    Nov 19 '19 at 10:07

I think that you are correct in the hint that you're remembering that the exercise told you to consider.

My idea is that, using this hint, you ought to be able to proceed inductively in constructing the regular complex $Y$.

Start by looking at all attaching maps $\varphi_{\alpha}:S^{\alpha} \rightarrow X_0$ onto the $0$-skeleton for $X$. Since $X$ is finite, these can be listed: $\{\varphi_{\alpha_1},...,\varphi_{\alpha_r}\}$.

Now, as you suggest, for each of these we wish to somehow consider instead the map $S^{\alpha} \rightarrow X_0 \times D^{\alpha+1}$, because this would be injective. What we therefore do is, for each $\alpha_1,...,\alpha_r$, take the successive product of $X_0$ with each of the $D^{{\alpha_i}+1}$s. This results in a space

$$Y_0 := X_0 \times D^{{\alpha_1}+1} \times....\times D^{{\alpha_r}+1}.$$

We can now use the standard cellular decomposition of $D^{k+1}$ to obtain a cellular decomposition of the product above. This is the "bottom" of our new regular complex $Y$. It is not the $0$-skeleton for our CW-complex $Y$, but it is homotopy equivalent to the $0$-skeleton of $X$. As such, we do indeed mildly abuse notation by calling this thing $Y_0$.

We then proceed with attaching cells to this product via the maps $\psi_{\alpha}$ as you suggest above. Importantly, we proceed as we normally would when we construct a CW-complex. We first attach the $1$-cells, and procede with $2$-cells, etc.

It is important to note a couple of things at this point.

  • We have certainly dealt with all attaching maps for the $1$-skeleton of $X$ at this point. What this means is that this process can indeed now be iterated; we look at attaching maps onto $X_1$, take whatever products we need to take with disks D^{\alpha}$ in order that these maps will be injective, and continue.

  • Importantly what we have constructed (and what we construct inductively in subsequent steps) is homotopy equivalent to $X_0$, $X_1$ and so on. It is indeed easy enough to see that the product $X_0 \times D^{{\alpha_1}+1} \times....\times D^{{\alpha_r}+1}$ is homotopy equivalent to $X_0$, but what about the next steps? We wish to see that $Y_1$ is homotopy equivalent to $X_1$. This can be seen in a few different ways, but the simplest for these purposes is just to state the following lemma (which I state rather wordily):

Lemma: Due to the nature of the construction of CW-complexes, whatever space $X$ you're attaching a particular cell to can be replaced by a homotopy equivalent space $\overline{X}$ (and the attaching map by the appropriate homotopic map) and the space which results $\overline{X} \cup D^{k}$ is homotopy equivalent to $X \cup D^k$.

Now what we're essentially doing is replacing $X_0$ with the homotopy equivalent $Y_0$, and attaching each $1$-cell to this new thing. Utilising the above lemma for each cell we attach gives that the resulting space is homotopy equivalent to $X_1$. We now "fatten up" this resulting space in exactly the same manner as how we "fattened up" $X_0$, by taking products with disks. Taking products with disks trivially results in a homotopy equivalent space, and what we end up with is the space which we will label $Y_1$; homotopy equivalent to $X_1$.

I think that iteratively doing this should complete the proof.

I should note that whilst the $Y_0$ and $Y_1$ which I mention above are certainly not the $0$ or $1$ skeletons for the space $Y$, there is nevertheless a natural $CW$-complex structure available for $Y$ which is obtained by using the standard cellular decomposition of the disks $D^k$, and this is then indeed a regular complex.

  • $\begingroup$ To conclude that $Y$ is regular, we tacitly use the fact that $D^k$ is a regular cell complex, correct? $\endgroup$ Nov 25 '19 at 7:58
  • $\begingroup$ We do indeed. When taking the product/s with the aforementioned disks, we need the space which results to have a regular CW-structure. I presume that this is where you're using that fact? $\endgroup$
    – Matt
    Nov 25 '19 at 9:10

References showing that CW-complexes have tha same homotopy type as simplicial CW-complexes (polyhedra):

Mardešić, Sibe. Lecture Notes on Spaces Having the Homotopy Type of CW-complexes. Department of Mathematics, University of Kentucky, 1978.

Milnor, John. "On spaces having the homotopy type of a CW-complex." Transactions of the American Mathematical Society 90.2 (1959): 272-280.

Lundell, Albert T., and Stephen Weingram. The topology of CW complexes. Springer Science & Business Media, 2012.

The first reference shows that finite CW-complexes are homotopy equivalent to finite polyhedra of the same dimension.


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