Given a $0$-connected $H$-space $X$, we can prove that left and right multiplication are (weak) homotopy equivalences. This allows us to construct the Hopf fibrations for $S^1$, $S^3$ and $S^7$. I always thought that $S^0$ was the odd one out due to it not being connected however after reading page 4 of "H-spaces from a homotopy point of view" I see that it is claimed $0$-connectedness can be weakened to $\pi_0(X)$ being a group.

Does this mean the real Hopf fibration can be constructed using the Hopf construction?

If so could I have some modern references?

This question originally stemmed from the discussion here on the nforum.


1 Answer 1


This is completely elementary. From the H-space structure $X\times X\to X$ you construct the Hopf fibration $X*X\to SX$ via $(x,t,y)\to (xy,t)$, where you think of $SX$ as the quotient of $X\times I$ by identifying both the $t=1$- and $t=-1$-level to one point, respectively. For $X=S^0$ you get a fibration $S^1\to S^1$, which you „see“ is the map of degree $2$: each point has exactly two preimages. So it is actually the fibration $S^1\to RP^1$.

  • $\begingroup$ What are the necessary conditions for $X*X \to SX$ to be a fibration? I seem to recall X being path connected is used some essential way. $\endgroup$ Feb 11, 2019 at 18:26
  • 1
    $\begingroup$ For $X=S^0$ you just see ad hoc what the map is: it is a 2-fold covering. $\endgroup$
    – ThiKu
    Feb 11, 2019 at 19:09

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