Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is $\mathbb{R}^n$ homeomorphic to a product $X \times Y$ with $X$ compact and not a point?

Bing's Dogbone space is a quotient of $\mathbb{R}^3$ with fibers points and arcs, and whose product with $\mathbb{R}$ is $\mathbb{R}^4$, so it doesn't seem to me to big a stretch to think that it may be possible.

Or, is there a notion of dimension which takes care of it swiftly?

share|improve this question
Well, X and Y must be contractible, so trivial cohomology. Y minus a point should then have the same cohomology as $S^{n-1}$. I think this imples that $\{x\}\times Y$ is open, so that $X$ is a discrete space, giving a contradiction. This seems like it should work... –  George Lowther Dec 16 '11 at 23:00
@George : Is it clear that $X$ and $Y$ must be contractible? Certainly that holds if they are homotopy equivalent to CW complexes, but I don't see how to deduce this in the general case. –  Andy Putman Dec 16 '11 at 23:03
Compare with math.stackexchange.com/questions/77175/… where the same question was asked for $S^n$. –  Alain Valette Dec 17 '11 at 6:26
For $S^n$ it can be done more quickly. Suppose that $X\times Y=Z$ for a non-contractible space Z. If Z minus a point is contractible, then X,Y are contractible so Z is contractible (contradiction), or one of X,Y is a single point. –  George Lowther Dec 17 '11 at 9:59
@George: that's a cool proof! –  Alain Valette Dec 18 '11 at 15:48

3 Answers 3

up vote 36 down vote accepted

No it is not possible. Suppose that $X\times Y\cong\mathbb{R}^n$. Then, as the product is contractible, both $X$ and $Y$ must be contractible spaces. For any $x\in X$, I'll show that $\lbrace x\rbrace\times Y$ must be an open subset of $\mathbb{R}^n$, which will imply that $\lbrace x\rbrace$ is an open subset of $X$ and, hence, that $X$ is discrete. Discrete contractible spaces consist of a single point.

Choose any $p=(x,y)\in X\times Y$. We just need to show that this is contained in the interior of $\lbrace x\rbrace\times Y$. As the spaces are contractible, there are deformation retractions $H_X\colon X\times[0,1]\to X$ and $H_Y\colon Y\times[0,1]\to Y$ respectively to the points $x,y$. So, $H_X(u,0)=u$, $H_X(u,1)=x$, $H_Y(v,0)=v$, $H_Y(v,1)=y$, for any $u\in X$ and $v\in Y$. Define the deformation retraction $J\colon(X\times Y)\times[0,1]\to X\times Y$ from $X\times Y$ to the point $p=(x,y)$ by $$ J\left((u,v),t\right)=\begin{cases} \left(H_X(u,2t),v\right),&\textrm{if }t\le1/2,\cr \left(x,H_Y(v,2t-1)\right),&\textrm{if }t\ge1/2. \end{cases} $$

Identifying $X\times Y$ with $\mathbb{R}^n$, consider the (n-1)-sphere $S_R=\lbrace a\in\mathbb{R}^n\colon\Vert a-p\Vert=R\rbrace$, for any fixed $R > 0$. As $K=X\times\lbrace y\rbrace$ is compact, it will have empty intersection with $S_R$ so long as $R$ is chosen large enough. However, retricted to $S_R\times[0,1]$, $J$ continuously deforms $S_R$ down to the single point $\lbrace p\rbrace$. This implies that $J(S_R\times[0,1])$ contains the open ball of radius $R$ centered at $p$. As $S_R\cap K=\emptyset$, $J(S_R\times[0,1/2])$ is a compact set not containing $p$. So, $J(S_R\times[1/2,1])\subset\lbrace x\rbrace\times Y$ contains a neighborhood of $p$, showing that $\lbrace x\rbrace\times Y$ is open in $\mathbb{R}^n$.

share|improve this answer
That's very nice! Thanks! –  Richard Kent Dec 17 '11 at 0:15

Here is a proof which uses only singular homology.$\newcommand{\RR}{\mathbb{R}}$$\newcommand{\ZZ}{\mathbb{Z}}$$\newcommand{\To}{\longrightarrow}$$\def\set#1{\lbrace#1\rbrace}$$\newcommand{\Xminusx}{X\setminus\set{x}}$$\newcommand{\Yminusy}{Y\setminus\set{y}}$

Assume $f:X\times Y\to\RR^n$ is a homeomorphism, and that $X$ is compact. I will prove that $X$ is a singleton by applying repeatedly the Künneth theorem, and using a few basic calculations of singular homology. By default, I use homology with coefficients in $\ZZ$. One can also carry out the exact same proof using homology with coefficients in a field, but the resulting simplifications are fairly inconsequential.

General remarks

The spaces $X$ and $Y$ cannot be empty, and we will fix $x\in X$ and $y\in Y$. Let also $p=f(x,y)\in\RR^n$. Observe that $X$ and $Y$ are Hausdorff, given that $\RR^n$ is Hausdorff. In particular, $\Xminusx$ is open in $X$, and $\Yminusy$ is open in $Y$. Furthermore, as observed in the comments, $X$ and $Y$ are contractible since $\RR^n$ is contractible. In particular, $H_\ast(X)$ is zero in positive degrees, and is $\ZZ$ in degree zero.

Claim 1: $H_n(Y,\Yminusy) \simeq H_n\bigl(\RR^n,\RR^n\setminus f(X\times\set{y})\bigr)$

First of all, consider the pair $X\times(Y,\Yminusy)=\bigl(X\times Y,X\times(\Yminusy)\bigr)$. By the Künneth theorem and the contractibility of $X$, we conclude that $$ H_n\bigl(X\times Y,X\times (\Yminusy)\bigr) = H_0(X)\otimes H_n(Y,\Yminusy) = H_n(Y,\Yminusy) $$ The above pair $\bigl(X\times Y,X\times (\Yminusy)\bigr)$ is homeomorphic via $f$ to $\bigl(\RR^n,\RR^n\setminus f(X\times\set{y})\bigr)$. The preceding expression thus implies $$ H_n\bigl(\RR^n,\RR^n\setminus f(X\times\set{y})\bigr) \simeq H_n(Y,\Yminusy) $$

Claim 2: $H_n(Y,\Yminusy)$ has a $\ZZ$ summand

Since $X$ is compact, the image $f(X\times\set{y})$ is compact in $\RR^n$, and thus bounded. Let $R\in\RR^+$ be such that $f(X\times\set{y})$ is contained in the closed ball of radius $R$ centered at $p$, $B_R(p)$. Then we have inclusions $$ \RR^n\setminus B_R(p) \subset \RR^n\setminus f(X\times\set{y}) \subset \RR^n\setminus\set{p} $$ which induce homomorphisms on homology: $$ \ZZ \simeq H_n(\RR^n,\RR^n\setminus\set{p}) \To H_n\bigl(\RR^n,\RR^n\setminus f(X\times\set{y})\bigr) \To H_n\bigl(\RR^n,\RR^n\setminus B_R(p)\bigr) \simeq \ZZ $$ The composition of the two maps is an isomorphism, therefore they exhibit a splitting of the middle group: $$ H_n(Y,\Yminusy) \simeq H_n\bigl(\RR^n,\RR^n\setminus f(X\times\set{y})\bigr) \simeq \ZZ \oplus A $$ for some abelian group $A$.

Claim 3: $H_\ast(X,\Xminusx)$ is concentrated in degree zero

Let $i$ be a positive integer. Observe that $f$ gives a homeomorphism between the pairs $\bigl(X\times Y,(X\times Y)\setminus\set{(x,y)}\bigr)$ and $(\RR^n,\RR^n\setminus\set{p})$. Consequently, $$ H_{n+i}\bigl(X\times Y,(X\times Y)\setminus\set{(x,y)}\bigr) \simeq H_{n+i}(\RR^n,\RR^n\setminus\set{p}) = 0 $$

Recall that $\Xminusx$ and $\Yminusy$ are open in $X$ and $Y$, respectively. So we can apply the Künneth theorem to the pair $$ (X,\Xminusx)\times(Y,\Yminusy) = \bigl(X\times Y,(X\times Y)\setminus\set{(x,y)}\bigr) $$ which implies that there is a monomorphism $$ H_i(X,\Xminusx)\otimes H_n(Y,\Yminusy) \To H_{n+i}\bigl(X\times Y,(X\times Y)\setminus\set{(x,y)}\bigr) = 0 $$ It follows that $H_i(X,\Xminusx)\otimes H_n(Y,\Yminusy) = 0$. Since $H_n(Y,\Yminusy)$ contains a summand isomorphic to $\ZZ$, we conclude that $H_i(X,\Xminusx)=0$.

Claim 4: $H_0(X,\Xminusx)$ is not zero

We now know that $H_\ast(X,\Xminusx)$ is zero in positive degrees, and it is necessarily a free abelian group in degree zero. Applying once more the Kunneth theorem to $(X,\Xminusx)\times(Y,\Yminusy)$, we obtain an isomorphism $$\begin{array}{rl} H_0(X,\Xminusx)\otimes H_n(Y,\Yminusy) \!\!\!\! & = H_n\bigl(X\times Y,(X\times Y)\setminus\set{(x,y)}\bigr) \\ & \simeq H_n(\RR^n,\RR^n\setminus\set{p}) \\ & \simeq \ZZ \end{array}$$ Consequently, $H_0(X,\Xminusx) \neq 0$.


Since $H_0(X)=\ZZ$, the only way that we can have $H_0(X,\Xminusx) \neq 0$ is if $\Xminusx = \emptyset$. Thus $X=\set{x}$ is a singleton.

share|improve this answer

This does not quite answer the question, but a related question (the title tells all you need to know):

Toruńczyk, H. Compact absolute retracts as factors of the Hilbert space. Fund. Math. 83 (1973), no. 1, 75–84.

share|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.