Let's consider two topological spaces. If they are homeomorphic, or homotopic equivalent, they have isomorphic fundamental groups, but the converse is not true. My question is: is there a (non trivial) equivalence relation between topological spaces that holds if and only they have isomorphic fundamental groups? Or weaker: what common properties have two spaces if they have isomorphic fundamental groups?

  • 2
    $\begingroup$ You can trivially define an equivalence relation for pathwise connected spaces by $X \sim Y$ if $\pi_1(X,x_0),\pi_1(Y,y_0)$ are isomorphic for some (or equivalently any) basepoints $x_0,y_0$. Is your question whether there exists an alternative definition of $\sim$ not using fundamental groups? $\endgroup$ – Paul Frost Feb 16 at 16:00
  • 1
    $\begingroup$ If they're sufficiently nice, their categories of covering spaces are equivalent; but I don't know if that's the sort of thing you're after $\endgroup$ – Max Feb 16 at 17:07
  • $\begingroup$ @PaulFrost some or any is not equivalent, in general. Basepoints matter. $\endgroup$ – Henno Brandsma Feb 16 at 17:09
  • $\begingroup$ @PaulFrost I wrote "non trivial"! $\endgroup$ – Marco All-in Nervo Feb 16 at 17:46
  • 3
    $\begingroup$ If they have no other non-trivial homotopy groups then $\pi_1$ contains essentially all their information. I don't think there is any other case where anything much can really be said. $\endgroup$ – Tyrone Feb 16 at 18:14

There is not really anything interesting that can be said. The property of two spaces having isomorphic fundamental groups is pretty much never singled out in the literature, because it doesn't have any interesting nontrivial consequences and doesn't arise naturally very often.

In some very special cases, there are stronger consequences. For instance, if $X$ and $Y$ are connected CW-complexes such that $\pi_n(X)$ and $\pi_n(Y)$ are trivial for all $n>1$, then $\pi_1(X)\cong \pi_1(Y)$ implies $X$ and $Y$ are homotopy equivalent. Or, if $X$ and $Y$ are both connected closed surfaces, then $\pi_1(X)\cong \pi_1(Y)$ implies $X$ and $Y$ are homeomorphic (this follows from the classification of surfaces).

  • $\begingroup$ I think you can take one of your statements slightly further: if $X$ and $Y$ are connected CW-complexes and we have an isomorphism $\alpha\colon \pi_1 (X)\to \pi_1 (Y)$, then for any $Z$ where $\pi_n (Z) = 0$ for $n>1$ I think the isomorphism $\alpha$ induces a bijection $[X, Z] \cong [Y,Z]$ by obstruction theory. In particular $H^1(X;R) \cong H^1(Y;R)$ for any $R$, and if $G$ is a discrete group then $X$ and $Y$ have the same theory of $G$-bundles. $\endgroup$ – William Feb 17 at 3:18

One term that people use for this relation is "have the same $1$-type", because it is the $1$-dimensional case of having the same $n$-type.

Definition: Two connected spaces have the same $n$-type if there is a zig-zag of continuous functions $X\leftarrow \dots \rightarrow Y$ which induce isomorphisms on homotopy groups up to degree $n$.

In the case $n = 1$, given abstract isomorphisms $\pi_1 X \cong G \cong \pi_1 Y$ they induce a zigzag

$$ X\to K(G, 1) \leftarrow Y$$

witnessing the fact that $X$ and $Y$ have the same $1$-type. In higher dimensions it's not enough to just have a sequence of abstract isomorphisms.

Below I'll outline an obstruction theory argument to prove the following (because I don't have a citation on hand):

Claim: If $X$ and $Y$ are CW complexes with the same $n$-type and $Z$ is any CW complex with $\pi_i(Z) = 0$ for $i > n$, then there is a bijection $[X, Z] \cong [Y, Z]$. (Here $[-,-]$ is the set of based-homotopy classes of maps.)

If $n = 1$, $Z$ satisfies this condition iff it is a $K(\pi_1(Z), 1)$, and so if $\pi_1(Z)$ is an abelian group $A$ this implies $H^1(Y;A)\cong H^1(X;A)$, i.e. if $X$ and $Y$ have isomorphic fundamental groups then they have isomorphic first singular cohomology groups with any coefficients (though there are other ways of seeing this which don't invoke Brown Representability). If $G$ is any discrete group then $BG \simeq K(G, 1)$, so $X$ and $Y$ will also have the same theory of principal $G$-bundles.

Obstruction Theory Argument

(I thought this discussion was in Hatcher somewhere, but I checked and couldn't find it. Another citation would be much appreciated because I have likely made mistakes.)

Let $X$ be a $CW$ complex. Let $X\lceil n \rceil$ be the complex obtained from $X$ by attaching cells of dimension $>n+1$ to kill all the homotopy groups $\pi_i(X)$ for $i > n$. (In particular $X$ and $X\lceil n \rceil$ share the same $(n+1)$-skeleton.) This space is sometimes called the $n$-th Postnikov stage of $X$.

For any spaces $X, Y, Z$, a continuous map $f\colon X \to Y$ defines a function $f_Z^*\colon [Y, Z] \to [X, Z]$ via pre-composition.

Lemma: If $\pi_i(Z)=0$ for all $i>n$ then the inclusion $\iota\colon X\to X\lceil n \rceil$ induces a bijection

$$ \iota_Z^*\colon [X\lceil n \rceil, Z] \cong [X, Z]$$

Proof sketch: We just need to argue that $i_Z^*$ is injective and surjective. For surjectivity suppose $f\colon X\to Z$; then $f$ is defined on the $(n+1)$-skeleton $X^{(n+1)}= X\lceil n \rceil^{(n+1)}$, and the obstructions to extending $f|_{X\lceil n \rceil^{(n+1)}}$ to some $\tilde{f} \colon X\lceil n \rceil\to Z$ live in the groups

$$ H^{r+1}(X\lceil n \rceil, X\lceil n \rceil^{(n+1)};\pi_r Z) $$

which vanish by assumption on $Z$. To see that $f$ is homotopic to $\tilde{f}\circ\iota$, note that they already agree on the $(n+1)$-skeleton and the rest of the obstructions live in

$$ \tilde{H}^r\big(X\lceil n \rceil/X\lceil n \rceil^{(n+1)};\pi_r Z\big)$$

so again they all vanish. For injectivity we need to consider two maps $f, g\colon X\lceil n \rceil\to Z$ and suppose there is a homotopy $H\colon f\circ \iota \sim g\circ \iota$. By cellular approximation we can assume the homotopy restricts to a homotopy of their restrictions to the $(n+1)$-skeleton. Then, similar to above, the obstructions to extending this restricted homotopy to a homotopy $f\sim g$ live in groups that vanish.

Proof sketch of the Claim: If two spaces $X$ and $Y$ have the same $n$-type via a zig-zag of functions, then by a bit more obstruction theorey and Whitehead's Theorem the zig-zag induces a homotopy equivalence $X\lceil n \rceil \simeq Y\lceil n \rceil$. Therefore by the above lemma

$$ [Y, Z] \cong [Y\lceil n \rceil, Z] \cong [X\lceil n \rceil, Z] \cong [X, Z]. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.