# Does Euclidean space have a compact factor?

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?

• 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

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 $\mathbb{R}^n$.

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,$H_0(X,\Xminusx) \neq 0$. ### Conclusion 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.

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.