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A wedge sum of circles without the gluing point is not path connected

I know that figure eight is not a manifold because its center has no neighborhood homeomorphic to $\mathbb{R}^n$. But how to prove this strictly?

Community
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Gobi
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1 Answers1

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Suppose that there was a neighborhood $U$ of the center point $P$ that was homeomorphic to $\mathbb{R}^n$. Consider $U \setminus \{P\}$. How many connected components does it have? How many connected components are there in $\mathbb{R}^n \setminus \{\text{point}\}$? [Be careful to note that the answer is different for $n=1$ than for $n > 1$, but that doesn't ultimately cause any trouble.]

Michael Joyce
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  • I got it, but can I assume that $U$ is a small nbd of $P$? Then the connected components will be 4, but if $U$ is strangely shaped, the components can be more than 4. – Gobi Oct 22 '12 at 12:55
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    @Gobi: $P$ has arbitrarily small $+$-shaped nbhds; no point of $\Bbb R^n$ has such a nbhd. See this answer and the comments following it. – Brian M. Scott Oct 22 '12 at 12:58
  • Sorry for the necropost. I think is safe to assume if the figure 8 was a manifold, it would be a 1-manifold. But if you remove P, I see just two components: The "North " one and its "South" counterpart. Just as you'd find two connected components in $\mathbb R - {p} $ – MSIS Feb 04 '22 at 23:46