In general, $\pi_0(X)$ is the set of path components of $X$ and does not have a group structure. After all, $S_0$ is just two points and the usual way of multiplying using the equation of a sphere doesn't work. But sometimes it is. For example, if $X$ is an H space we still have a 0-th homotopy group.

What are other cases? In particular, if $G$ is a discrete group (group with the discrete topology), then is $\pi_0(G)=G$ as a set and as a group?

  • $\begingroup$ What is your opinion? $\endgroup$
    – user122283
    Mar 29, 2014 at 23:09
  • 1
    $\begingroup$ I just realized you mentioned H-spaces in your question. You... realize every group is an H-space, right? So a discrete group trivially has that property. $\endgroup$
    – user98602
    Mar 29, 2014 at 23:36
  • $\begingroup$ I was aware of the statement for H-spaces. But I was under the impression that H-spaces were required to be path-connected spaces. $\endgroup$
    – Josh
    Mar 30, 2014 at 19:41
  • $\begingroup$ (I just realized that sentence sounded rude - that wasn't intended at all! Sorry!) I wasn't under the impression that an H-space needed to be path-connected; in that case, $\pi_0$ is of course the trivial group. (I'm not aware of the proof that an H-space has $\pi_0$ a group. I think you're right that we need something stronger than H-space in that case, if we allow H-spaces to be disconnected.) $\endgroup$
    – user98602
    Mar 30, 2014 at 19:49

1 Answer 1


When $G$ is a topological group, we always have that $\pi_0(G)$ is a (topological) group!

Lemma: The connected component of the identity in $G$ - let's denote it $C$ - is a normal subgroup of $G$. By continuity of the group operation, $f(C,C)$ is connected. It contains the identity, so $f(C,C) \subset C$ because $C$ is the connected component of the identity. Similarly, $C^{-1} \subset C$ by continuity of the inversion operator. So $C$ is a subgroup.

We need to show normality still: that $xCx^{-1} \subset C$ for $x \in G$. Continuity (again!) shows us that $f(xC, x^{-1})$ is connected, because $xC = f(x,C)$ is connected, so that $xCx^{-1} \subset C$. This proves what we wanted to prove.

Now $G/C$ is the set of connected components of $G$, and is a group. So $\pi_0(G)$ always inherits a group structure from $G$.

  • $\begingroup$ The map f is the group operation? $\endgroup$
    – Josh
    Mar 30, 2014 at 19:43
  • $\begingroup$ Yes, I should have clarified. $f(g,h) = gh$. $\endgroup$
    – user98602
    Mar 30, 2014 at 19:44
  • $\begingroup$ So you mean $f(C,C)$ to mean $\{gh \in G | g, h \in C\}$? $\endgroup$
    – Josh
    Mar 30, 2014 at 19:47
  • $\begingroup$ @jd.r Yep (and similarly in other cases, where I make the arguments as sets rather than elements). $\endgroup$
    – user98602
    Mar 30, 2014 at 19:50
  • $\begingroup$ That's what I thought, and what made sense in context...just wanted to be sure. $\endgroup$
    – Josh
    Mar 30, 2014 at 20:14

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