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Is there any subset of $R^n$ homotopically equivalent to the wedge product of countable many circles. In particular, is the union of circles in $R^2$ with center (0,n) and radius n for n=1,2 ... n (sort of inverse Hawaiian earring) homotopically equivalent to that wedge product.

Note that the answer is no if we ask about topological equivalence: the wedge product is closed but not locally compact.

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Don't know, but it looks like the obvious bijection induces an isomorphism of fundamental groups. –  Robin Chapman Oct 24 '10 at 18:47
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The "inverse Hawaiian earring" is not locally compact. (No neighborhood of $(0,0)$ in it is closed in the plane.) But it is distinguishable from the infinite wedge of circles by the fact that the latter is not metrizable: the interesting point has no countable neighborhood base. –  Tom Goodwillie Oct 24 '10 at 21:30
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3 Answers 3

up vote 4 down vote accepted

You can certainly embed the real line, with perpendicular circles at the integer points, into $\mathbb{R}^3$, and this is homotopy equivalent to the wedge you want.

Let $W$ be the expanding `inverse Hawaiian earring'.
Then the obvious map $\bigvee S^1\to W$ is a homotopy equivalence. To define the inverse, let $Z$ be the intersection of $W$ with a small closed rectangle around the origin;
collapse $Z$ to a point, then map the resulting quotient back to $\bigvee S^1$ by the inverse of the obvious map.

We have to check that $W\to \bigvee S^1$ is continuous.
If $U\subseteq \bigvee S^1$ is open and does not contain the basepoint $\star$, then the preimage is obviously open. If $\star\in U$, then the preimage is the preimage of $U- \star$ (which is open) together with a neighborhood of $Z$, which is also open.

Since $Z$ is contractible and its inclusion into $W$ is a cofibration, the composite $W \to \bigvee S^1\to W$ is homotopic to $\mathrm{id}_{W}$. In essentially the same way, the composite $\bigvee S^1\to W\to \bigvee S^1$ is homotopic to the identity.

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Take $\mathbb{R}^2$ and remove the points $(n,0)$ for $n\in \mathbb{Z}$ along the $x$-axis. This is homotopically equivalent to the wedge of countably many circles. Take the union of a (clockwise) loop around every deleted point, based at (0,-1), and chosen to be disjoint from each other. This is the image of the countable wedge and the described space deformation retracts onto it, making them homotopically equivalent.

Alternatively, delete only the points at positive integers along the $x$-axis. Then take loops based at $(-1,0)$ which are circles of increasing radius, including one more deleted point for each new circle (this is a sort of reflected Hawaiian earring). The inclusion of the subspace which is the union of all the circles is a homotopy equivalence between the countable wedge and this new space.

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The infinite wedge product of $S^1$ is a CW complex, so it is not homeomorphic to the inverse Hawaian earing. Is this correct?

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Your remark is correct, but unhelpful. –  André Henriques Sep 10 '11 at 16:29
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