# 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?

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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

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 $\mathbb{R}^n$.

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That's very nice! Thanks! –  Richard Kent Dec 17 '11 at 0:15


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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.

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