I have to prove that two different topological spaces $X,Y$ have the same homotopy type. I've been able to prove so far that $\pi_1(X)=\pi_1(Y)$ but I don't know if this is enough to say that $X$ and $Y$ are isomorphic in the category $\mathcal{HoTop}$.

I also know that two spaces with the same fundamental group do not have to be homeomorphic (i.e. an isomorphism in the category $\mathcal{Groups}$ does not imply that the objects are isomorphic in the category $\mathcal{Top}$), but I was wondering what is the situation in $\mathcal{HoTop}$. Are two spaces with the same fundamental group homotopy equivalent?

I don't know if the notation that I'm using is usual or not, so if anyone needs clarification do not hesitate of commenting.

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    $\begingroup$ Both a point and the two sphere have trivial fundamental groups. $\endgroup$ – lulu Nov 10 '15 at 19:40
  • $\begingroup$ The plane $\mathbb{R}^2$ and the 2-sphere $S^2$ have trivial fundamental groups, but $S^2$ is compact, while the plane is not, so... they're not homotopy-equivalent. $\endgroup$ – BrianO Nov 10 '15 at 19:49
  • $\begingroup$ @DanielGerigk That's much easier anyway. I thought to delete my bogus comment, but I'll leave it — it's informative that compactness isn't homotopy-invariant. $\endgroup$ – BrianO Nov 11 '15 at 2:37
  • $\begingroup$ @BrianO Compactness is not homotopy invariant ($\mathbb{R}^2\simeq \{*\}$). $\endgroup$ – Nitrogen Nov 11 '15 at 6:17
  • $\begingroup$ @Nitrogen Yyyyep. Of course \{\}$ *is compact, so... big "doh" on my part. $\endgroup$ – BrianO Nov 11 '15 at 6:19

No. As lulu already pointed in a comment, a point and $\mathbb{S}^2$ have the same fundamental group but are not homotopically equivalent ($\mathbb{S}^2$ is not contractile, say).

There is a result, the Whitehead Theorem, which states that if $X$ and $Y$ have CW complex structures, and $f\colon X \to Y$ is a map such that $f_\ast\colon \pi_n(X,x_0) \to \pi_n(Y,f(x_0))$ is an isomorphism for all $n \geq 1$, then $X$ and $Y$ have the same homotopy type.

Here $\pi_n(X,x_0)$ is the space of homotopy equivalence classes of hyperloops $\alpha: \mathbb{S}^n \to X$ such that $\alpha(1,0,\cdots,0) = x_0$.

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    $\begingroup$ For the Whitehead Theorem, we also need the assumption that $f_0 : \pi_0(X,x_0) \rightarrow \pi_0(Y,f(x_0))$ is an isomorphism. $\endgroup$ – Daniel Gerigk Nov 11 '15 at 12:19

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