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In exercise 8.2 of Rotmans "An Introduction to Algebraic Topology", we have to show $\mathbb{R}P^1\cong S^1$, and in exercise 8.5 that for all $n\geq0$ we have $\mathbb{R}P^n\cong S^n/\sim$ where $\sim$ identifies antipodal points. So, does this imply $S^1\cong S^1/\sim$? This seems strange to me and I would be glad if someone could clarify this for me. Thanks in advance.

Thanks.
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    I just want to say that this shouldn't seem too strange, come up with some other spaces that have quotients homeomorphic to the original space: fold a square in half, for example. – Justin Young Sep 28 '20 at 15:54
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    Note also that if you pick, say, a Riemannian metric on the circle then the quotient circle has half the length of the original. – Qiaochu Yuan Sep 28 '20 at 16:23
  • @JustinYoung I think, "strangeness" is strongly related to experience, and for me as a beginner in topology there are a lot of strange things which shouldn't be that strange - thanks for the remark, it did a great job clarifying one of them. But from an algebraic point of view, objects (say groups) being isomorphic to some quotient of themselves is really more unnatural? I don't think I have seen an example in that direction (trivial quotients excluded, of course.) – Thanks. Sep 28 '20 at 19:27
  • @Const: $S^1$ is actually an example of a group with a proper quotient isomorphic to itself also. – Qiaochu Yuan Sep 29 '20 at 01:52
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    Just to clarify, I didn't mean to chastise you for finding it strange, I just meant that if you encounter a variety of examples in topology, you will see this can come up easily. This is how learning happens. There are many groups isomorphic to proper quotients (infinite groups of course) of themselves. The simplest example I can thank of would be $\bigoplus_{i=0}^\infty \mathbb Z$ (which you could view as $\mathbb Z[x]$). – Justin Young Sep 29 '20 at 02:38
  • @JustinYoung Sorry for the sort of misunderstanding, I meant it by now ways negative - I just thought it‘s like a child growing up and the younger you are, the more surprises you, which might also be the beauty of learning. Thanks for this remark, it helped me clarifying this: I just need to take a surjective self-homomorphism, say in $\mathbb{Z}[x]$ the left shift? – Thanks. Sep 29 '20 at 07:10
  • Thanks, @QiaochuYuan, so for $S^1$ one could wrap it around itself two-times and then pass to the quotient modulo it‘s kernel? – Thanks. Sep 29 '20 at 07:12
  • Are there any sort of criteria for a group having a surjective self-morphism with non-trivial kernel? Both examples above seem for me to have some sort of restriction involved, for in $\mathbb{Z}[x]$ we can perform a shift and in $S^1$ we can „speed it up continuoesly“? – Thanks. Sep 29 '20 at 07:37
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    @Const: the term is "non-Hopfian" (https://en.wikipedia.org/wiki/Hopfian_group); that is, a Hopfian group is a group not isomorphic to any of its proper quotients. Lots of theorems are known about which groups are Hopfian or non-Hopfian. – Qiaochu Yuan Sep 29 '20 at 17:34

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Yes, that's right, the circle is homeomorphic to itself mod antipodal points. One way to think about this is to think about the circle as the interval $[0, 1]$ with the endpoints glued. When you identify antipodal points, this is equivalent to gluing $a$ to $(a + \frac{1}{2}) \text{ mod } 1$. The resulting space is clearly $[0, \frac{1}{2}]$ with the endpoints glued, which is just another circle topologically.

hunter
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