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Can we prove this by merely using point-set topology, without mentioning algebraic topology concepts?

Or do I have to know of basic algebraic topology such as fundamental group and etc..?

Anyway please give some simple proof for this one.

And one more thing. I have Munkres's Topology textbook. Which part of it should I read to prove something like this? I've read most of the part in the 'General Topology'

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    $\begingroup$ Hint: Will the figure-eight stay connected if you remove a certain point? $\endgroup$ Commented Sep 24, 2013 at 12:32
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    $\begingroup$ Point-set topology is enough. What happens to connectedness if you remove one point? $\endgroup$ Commented Sep 24, 2013 at 12:33
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    $\begingroup$ Oh thanks guys. It was an easy one! $\endgroup$
    – le4m
    Commented Sep 24, 2013 at 12:38
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    $\begingroup$ @julypraise: If you want to write up a formal proof as an answer, it will cut down on the number of unanswered questions on the site. You can accept your own answer after a few days (three, I think), and answering your own question is explicitly encouraged! $\endgroup$ Commented Sep 24, 2013 at 13:26

2 Answers 2

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For the sake of lowering the number of unanswered questions:

If there was a homeomorphism between both figures, say $\;f: S^1\vee S^1\to S^1\;$ , then also $\;\overline f: \left(S^1\vee S^1\right)\setminus\;\{x_0\}\to S^1\setminus\{f(x_0)\}$ , would be a homeomorphism (why?).

But if $\,x_0=$ the intersection (in fact, the tangent) point of the two circles, then $\;\left(S^1\vee S^1\right)\setminus\{x_0\}\;$ is not connected, yet $\;S^1\setminus\{f(x_0)\}\;$ still is connected, no matter what $\;f(x_0)\in S^1\;$ is.

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We argue by reasoning about path-connectedness in each shape. For the figure 8 shape, picking any two distinct points---one in the left half of the shape and one in the right half, we observe that all paths connecting these two points must share one common point (i.e. the point at the center of the shape). Suppose there exists a homeomorphism $f$ from figure 8 to $\mathbb{S}^1$. Using $f$ to map the paths connecting the two points in figure 8 to $\mathbb{S}^1$, we have that all paths connecting the image of the two points in $\mathbb{S}^1$ cross a common point in $\mathbb{S}^1$ as well, which is false---there always exists (completely) distinct paths in $\mathbb{S}^1$ connecting two distinct points. Hence, the figure eight figure 8 cannot be homeomorphic to the circle $\mathbb{S}^1$.

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