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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
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Compare with math.stackexchange.com/questions/77175/… where the same question was asked for $S^n$. –  Alain Valette Dec 17 '11 at 6:26
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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
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@George: that's a cool proof! –  Alain Valette Dec 18 '11 at 15:48
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3 Answers

up vote 34 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 $\mathbb{R}^n$.

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

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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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