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I know that if two spaces are path-connected and homotopy equivalent then their fundamental groups are isomorphic Does this go the other way so we can say that if two spaces have isomoprhic fundamental groups then they are homotopy equivalent?

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  • $\begingroup$ By that, do you mean just $\pi_1(X) \simeq \pi_1(Y)$, or that $\pi_n(X) \simeq \pi_n(Y)$ for each $n \ge 1$? $\endgroup$ Commented Apr 6, 2021 at 22:57
  • $\begingroup$ Just the first homotopy group not the higher ones $\endgroup$ Commented Apr 6, 2021 at 22:58
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    $\begingroup$ no. for instance, the $2$-sphere has trivial fundamental group, but it is not homotopy equivalent to a point. (this is not trivial; if you would like a reference let me know.) $\endgroup$ Commented Apr 6, 2021 at 23:01
  • $\begingroup$ Thank you, this puts things into perspective $\endgroup$ Commented Apr 7, 2021 at 21:25
  • $\begingroup$ @AtticusStonestrom might you consider answering your comment so that an answer can be accepted. $\endgroup$
    – Anthony
    Commented Apr 7, 2021 at 23:36

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No. Consider two spheres $S^n$ and $S^m$ with $n,m>1$. Then $\pi_1(S^n)=\pi_1(S^m)=0$ while $S^n$ is not homotopy equivalent to $S^m$ whenever $n\neq m$ (which can be seen by simple calculation of homology groups).

For more sophisticated example consider the double comb space $X$, which has all homotopy groups trivial $\pi_n(X)=0$, homology groups trivial $H_n(X)=0$ and cohomology groups trivial $H^n(X)=0$, but it is not contractible (it is not homotopy equivalent to a point).

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  • $\begingroup$ Thanks so much for taking the time to answer, its clear to me now $\endgroup$ Commented Apr 7, 2021 at 21:25

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