# Does the Hopf construction work for $S^0$?

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.

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 $$S^1\to RP^1$$.
• 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. Feb 11, 2019 at 18:26
• For $X=S^0$ you just see ad hoc what the map is: it is a 2-fold covering. Feb 11, 2019 at 19:09