MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Adams, J.F. Infinite Loop Spaces Princ. Univ. Press. page 9 he states Alexander duality theorem

Theorem:[Alexander Duality] $$ H^r(X,G)=H_{n-r+1}(S^n-X,G)$$

for finite CW-complexes with a "nice embedding". That is to say, $S^n-X$ has a CW-complex $Y$ as deformation retract and $X$ is a deformation retract of $S^n-Y$.

My first question is the following: Do you know if every CW-complex admits a nice embedding into some $S^n$? if so, could you give a reference?

Thanks in advance.

share|cite|improve this question
You have a typo in the title. "stalbe" – David White Oct 17 '11 at 13:15
Every finite CW complex embeds into some $S^n$ as proved e.g. in Hatcher's book Corollary A.10. Now in your definition do you want $Y$ to be a finite CW complex? If this is not required, you could just take $Y=S^n-X$ which has a CW structure (being a smooth manifold), and then $X=S^n-Y$, so trivially $X$ is a deformation retract of $S^n-Y$. If $Y$ must be finite, then one has to work a bit harder. – Igor Belegradek Oct 17 '11 at 13:54
Igor, I wasn't aware of that argument in hatcher (A.10). Thanks for sharing it. It supersedes my answer below. – John Klein Oct 17 '11 at 22:21
up vote 7 down vote accepted

1) The answer is yes, at least up to homotopy. This can be found in Wall's paper:

Wall, C. T. C., Classification problems in differential topology---IV. Thickenings. Topology, 1966, 5, 73–94.

Wall argues this in a cell-by-cell way, making use of transversality.

2) Alternatively, if X is a finite simplicial complex, then one can find an actual embedding by general position arguments. See Rourke and Sanderson's book for this. Note it is classical that every finite CW complex is homotopy equivalent to a finite simplicial complex.

share|cite|improve this answer
@Tintin: it seems that you didn't read my comment above under the question (dated from October 17, 2011 at 22:21). – John Klein Sep 9 '15 at 10:06
Sorry John. On the contrary, I read it. Since I marked your response as the answer to my question I wanted to add your comment to your response in case someone comes here and just read your answer. I correct it now. By the way, thanks for the help :) – Tintin Sep 9 '15 at 10:10
As remarked by Klein, complete this answer with the second comment of the question by Igor Belegradek. The result is also proved in Hatcher's book Corollary A.10. Thanks for the help – Tintin Sep 9 '15 at 10:13

Let $X$ be a finite CW-complex. Then there is an embedding $X\to\mathbb{R}^n$ for some $n$. To show this it suffices to consider the case when $X$ is obtained from a CW-complex $Y\subset B^m\subset \mathbb{R}^m$ by attaching a cell: $X=Y\cup_f B^k$ where $B^m$ is the unit ball in $\mathbb{R}^m$ and $f:S^{k-1}\to Y$ is the attaching map.

Let us show that there is an embedding $i:B^m\sqcup B^k\to \mathbb{R}^n$ for some $n$ so that, given $x',x''\in B^m$ and $y',y''\in B^k$, the segments $[i(x'),i(y')]$ and $[i(x''),i(y'')]$ do not intersect unless $x=x''$ or $y'=y''$. Indeed, let's assume both balls live in some $\mathbb{R}^r$ and take $\mathbb{R}^n$ to be the space of polynomials $\mathbb{R}^r\to\mathbb{R}$ of sufficiently high degree (4 will do). Then for any 4 points in $\mathbb{R}^r$ the conditions that an element of $V$ are zero at those points are linearly independent. So for any $x\in B^m\sqcup B^k$ take $i(x)\in V^*$ to be the evaluation function $p\mapsto p(x)$.

Now set $X'$ to be the union of $i(Y), i(B^k)$ and all segments $[i(x),i(f(x))]$ for $x\in S^{k-1}$. there is a natural bijective continuous map $X\to X'$, which is a homeomorphism since $X$ is compact and $\mathbb{R}^n$ is Hausdorff.


  1. From this construction it is clear that one can construct an embedding of $X$ that is smooth on the interior of each cell. So by slightly modifying the proof of Whitney's embedding theorem we see that $X$ can be embedded in $\mathbb{R}^{2\dim X+1}$. However, the strong Whitney theorem does not hold. E.g. there are 2-polyhedra that can't be embedded in $\mathbb{R}^4$, see e.g. theorem 3 in this paper

  2. If $X$ is not assumed finite then I would guess that a necessary and sufficient condition for $X$ to be properly embeddable in some $\mathbb{R}^n$ is that $X$ should be locally finite and should have at most countably many cells.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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