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Let $ X = \mathbb{S}^1 \times \{ \pm 1 \} \subset \mathbb{R}^3 $. We define on $ X $ the equivalence relation generated by $ (x,-1) \sim (x,+1) $ if $ x \neq 1 $, and consider the quotient $ S := X / \sim $.

Using the Seifert van Kampen theorem I have proven that S has the same fundamental group two circumferences glued by one point $p$, let's call it $C$, and its fundamental group is $\pi_1(C; p) = \langle a,b \rangle$ . And I want to see if $S$ and $C$ have the same homotopy type. I dont know how to start. Maybe supossing there exist
$ f : S \rightarrow C $, $ g : C \rightarrow S $ such that $ g \circ f \cong 1_X $ y $ f \circ g \cong 1_Y $, then removing one point and getting a contradicction. Or maybe it is true.

Any idea is welcome

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    $\begingroup$ Note that the fundamental group of the wedge of two circles is the free group on two generators, while the fundamental group of the torus is the free abelian group on two generators. Those are different groups. $\endgroup$ Commented May 16 at 16:04
  • $\begingroup$ Thanks, I forgot the fundamental group of the torus was abelian. I'll edit the question $\endgroup$
    – Alejandro
    Commented May 16 at 16:07
  • $\begingroup$ That would not make $S$ homeomorphic to $S^1 \land S^1$, as any continuous map $S \to S^1 \land S^1$ must send $((1, 0), -1)$ and $((1, 0), 1)$ to the same point (since the latter is Hausdorff). $\endgroup$ Commented May 16 at 16:58
  • $\begingroup$ Sorry, I've edited the question. It is $(x,-1) \sim (x,+1)$ if $x \neq 1 \in \mathbb{S}^1$ $\endgroup$
    – Alejandro
    Commented May 16 at 17:28
  • $\begingroup$ I am not saying that S is homeomorphic to anything, just that the fundamental groups of $S$ and $C$ are isomorphic. But even though they are not homeomorphic, they migth have the same homotopy type. $\endgroup$
    – Alejandro
    Commented May 16 at 17:31

1 Answer 1

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Those spaces are not homotopy equivalent. In fact, no map $f:S\to C$ induces an isomorphism on $\pi_1$.

Any $f:S\to C$ map has to map both special points: $[1,-1]$ and $[1,+1]$ to the same point, because $C$ is Hausdorff. This means that $f$ factorizes through the quotient of $S$ that identifies those points. Meaning $f$ factorizes through $\mathbb{S}^1$. And therefore $\pi_1(f)$, which is a $\mathbb{Z}*\mathbb{Z}\to\mathbb{Z}*\mathbb{Z}$ map, factorizes through $\mathbb{Z}$. And so it cannot be an isomorphism.

Note that in the above we don't need to know anything about $C$, except that it is Hausdorff. Yes, this shows that $S$ is not homotopy equivalent to any Hausdorff space.

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  • $\begingroup$ This leaves the possibility of a map $C \to S$ inducing an isomorphism on $\pi_1$ (and indeed on all higher homotopy groups. $\endgroup$ Commented May 16 at 21:49
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    $\begingroup$ @NaïmFavier It's indeed true that the particular map $C\to S$ you expect to induce an isomorphism on $\pi_1$ is a weak homotopy equivalence. In this sense, $C$ and $S$ have the same (weak) homotopy type. $\endgroup$ Commented May 16 at 21:54
  • $\begingroup$ I wonder how one might prove that. Can you show that all higher homotopy groups of $S$ are trivial? $\endgroup$ Commented May 17 at 15:23
  • $\begingroup$ @NaïmFavier It's probably easier to start with the following: $\mathbb{R}$ is the universal cover of an open interval with double origin, in basically the same way that $\mathbb{R}$ is the universal cover of the circle. Now, if you have this picture in mind, then you can see via a similar picture that the universal cover of $S^1\vee S^1$ is also the universal cover of $S$. As such, all higher homotopy groups of $S$ vanish. Of course, there are multiple proof obligations here, and I'll gladly leave spelling out the details to someone who wants to turn this into an answer. $\endgroup$ Commented May 17 at 22:20
  • $\begingroup$ This question is somewhat relevant: mathoverflow.net/questions/25472/… $\endgroup$ Commented May 19 at 11:24

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