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I know that if $h : (X,x_0) \longrightarrow (Y,y_0)$ is a homeomorphism then that induces an isomorphism $h_{*} : \pi_1(X,x_0) \longrightarrow \pi_1(Y,y_0)$ defined by $$h_{*} ([f]) = [h \circ f].$$ Now my question is "Does the converse of this hold?" i.e. Suppose I have an isomorphism $h : \pi_1(X,x_0) \longrightarrow \pi_1(Y,y_0)$. Does that necessarily induce a homeomorphism $h^{*} : (X,x_0) \longrightarrow (Y,y_0)$? If it is true then can we go further i.e. can we also say that the isomorphism induced by $h^{*}$ is same as that of $h$? Please help me in understanding this concept.

Thank you very much.

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    $\begingroup$ Of course it doesn't hold. E.g. $\Bbb R^n$ would be homeomorphic to a one point space.. $\endgroup$ – Berci Aug 11 '18 at 17:19
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Definitely not. Among other things, this would imply that any two simply connected spaces are homeomorphic.

More generally, the fact that homeomorphisms between topological spaces induce isomorphisms of their fundamental groups shows that the fundamental group of a space is a topological invariant. If the converse were true, then we could completely characterize topological spaces by their fundamental groups.

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  • $\begingroup$ Yeah you are right. If my assumption was true then $\mathbb R$ and $\mathbb R^2$ were homeomorphic because they are both simply connected and hence they both have the trivial fundamental group which are isomorphic. Let $f$ be the corresponding homeomorphism and let $f(0)=a$ then the restriction of $f$ to $\Bbb R \setminus \{0 \}$ is again a homeomorphism from $\Bbb R \setminus \{0 \}$ onto $\Bbb R^2 \setminus \{a \}$ which is a contradiction, since $\Bbb R^2 \setminus \{a \}$ is path connected but $\Bbb R \setminus \{ 0\}$ is not. $\endgroup$ – Dbchatto67 Aug 11 '18 at 18:32
  • $\begingroup$ In fact $\Bbb R^n$ would then be homeomorphic to a space containing a single point since every space containing one point is trivially path connected and hence simply connected. Very silly question. Sorry for asking this question without thinking a bit more. $\endgroup$ – Dbchatto67 Aug 11 '18 at 18:40
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The answer is a big no.

If you study the fundamental group a bit more, you'll see that it is actually homotopy invariant: that is if $X,Y$ have the some homotopy type (and are path-connected to make the statement easier) then $\pi_1(X,x_0)\simeq \pi_1(Y,y_0)$. There are homotopy equivalent spaces that are not homeomorphic, such as $\mathbb{R}$ and a point $\{*\}$: at best we can hope to classify spaces up to homotopy equivalence with the fundamental group.

It turns out that an isomorphic $\pi_1$ is not even enough to determine that two spaces have the same homotopy type: indeed there are many other homotopy invariants, such as the higher homotopy groups $\pi_n$: it could very well be that $\pi_1(X)\simeq \pi_1(Y)$ but $\pi_n(X)\neq\pi_n(Y)$ for some $n>1$, and in such a situation, $X$ and $Y$ can't have the same homotopy type; and this situation happens.

But it gets "worse": it's possible that for each $n\geq 1$, $\pi_n(X,x_0)\simeq\pi_n(Y,y_0)$, without $X$ and $Y$ having the same homotopy type. And even worse, it's possible that there is a map $f:X\to Y$ that induces an isomorphism on each $\pi_n$ but isn't a homotopy equivalence !

But it gets better. This last statement is wrong if we restrict to (very) nice spaces, called CW-complexes (this is Whitehead's theorem -note that the isomorphisms have to be induced by some map to begin with): that's one of the reasons why algebraic topologists like to restrict to CW-complexes, these nice spaces are a good place to study homotopy theory and homotopy groups.

In summary: $\pi_1$ is not enough to classify homeomorphism, not even homotopy equivalence; even the sequence of all $\pi_n$'s is not enough for that purpose; but for nice spaces, the sequence of all $\pi_n$'s is enough (in a certain sense) to classify the homotopy type of these (nice) spaces

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