## Can H-space multiplication always be straightened so that mult.-by-id. is the identity on the nose?

In JP May's Concise Course in Algebraic Topology, on page 143 he says that the left- and right-multiplication-by-identity maps $\lambda:X\rightarrow X$ and $\rho:X\rightarrow X$ specify a map $X\vee X\rightarrow X$ that is homotopic to the codiagonal map. This doesn't seem obvious to me, because the homotopies $H^L:\lambda\simeq 1_X$ and $H^R:\rho\simeq 1_X$ need not agree on how they move $\lambda(e)=\rho(e)$ to $e$; a priori, the paths $H^L_t(e)=H^L(e,t)$ and $H^R_t(e)=H^R(e,t)$ need not be homotopic. My intuition is failing me, because the only H-spaces I can think of are Lie groups and loop spaces. In both of those cases the multiplication already satisfies $e\times e \mapsto e$, and in the former case already $e\times x = x = x\times e$, while in the latter case the obvious homotopies $H^L$ and $H^R$ fix $e$ for all $t$...

-
$\lambda$ and $\rho$ are not "multiplication maps", they are "multiplication by $e$ maps". – Mariano Suárez-Alvarez Apr 19 2010 at 4:29
You will find two $H$-space structure on infinite projective space here mathoverflow.net/questions/11117/… – Mariano Suárez-Alvarez Apr 19 2010 at 4:30
Finally, in all likelyhood, May is assuming that everything is relative to the basepoint $e$. – Mariano Suárez-Alvarez Apr 19 2010 at 4:35
#1: That's what I meant, thanks. #2: I was thinking of E-M spaces, too, but if I understand correctly those can be given H-space structures just by considering them as loop spaces. #3: Yes, I realized that after I posted. I'd be still interested to know if this condition is necessary, even though in practice it probably almost always holds. – Aaron Mazel-Gee Apr 19 2010 at 6:24
@Aaron, re #2: But the thing is, they are not loop spaces, so you cannot consider them to be E-M spaces! They are of the homotopy type of loop spaces, so you can transport the structure along an homotopy equivalence, but then you do not get strict units in general, and so on. – Mariano Suárez-Alvarez Apr 19 2010 at 7:32

There are three possible definitions of an H-space, according to whether the "identity" element is a strict left and right identity, or only an identity up to basepoint-preserving homotopy, or just up to a non-basepoint-preserving homotopy. The three notions turn out to be equivalent, assuming the space is nice enough that one can apply the homotopy extension property when needed. The proof of equivalence was left as an exercise in my book (Exercise 1 in section 3.C), but it's actually a rather tricky argument, not really fair for an exercise, so a couple weeks ago I wrote up the proof and posted it here:

http://www.math.cornell.edu/~hatcher/AT/ATsolution3C.1.pdf

It's a one-page argument. If anyone knows a simpler proof, or a reference for this fact in the literature, I'd be happy to hear about it.

-
 I see. This is a little more involved than I hoped the answer would be, but it's what I was looking for. Thanks! – Aaron Mazel-Gee Apr 19 2010 at 17:56

I think you may have to assume that X is well-pointed, i.e. the inclusion $\{e\} \to X$ is a (Hurewicz) cofibration. Does May assume this? If this is the case, then any map which maps the base point to the component of the base point can be homotoped to a base-point preserving map, and any two base-point preserving maps which are homotopic (but the homotopy does not preserve the base point) are also homotopic via a base-point preserving homotopy.

For the first claim, you consider the problem of lifting $X \times \{0\} \cup \{x_0\} \times [0,1] \to X$ to $X \times [0,1] \to X$, where the first map consists of the original map and a path from the basepoint to $e$.

For the second claim, the suitable lifting problem is given by lifting

$$(\{x_0\} \times [0,1] \times [0,1]) \cup (X \times \{0,1\} \times [0,1]) \cup (X \times [0,1] \times \{0\}) \to Y$$

to $X \times [0,1] \times [0,1]$, where the original homotopy lives on the $X \times [0,1]$ factor. I can try to provide more detail if this isn't clear enough.

-
How do I get curly braces? \{...\} doesn't seem to work. – Tilman Apr 19 2010 at 6:49
I don't think May assumes this, no -- his term is "nondegenerately based", and he doesn't make that assumption until the first problem itself, where he asks you to obtain a map X x X --> X which agrees with the codiagonal on X v X. – Aaron Mazel-Gee Apr 19 2010 at 6:59
OK... well-pointed and nondegenerately based is the same thing. I think you need that here, too, but in any case, it's probably not worth the effort to think very hard about ill-based (degenerately-based) H-spaces. – Tilman Apr 19 2010 at 7:07
(Braces: either double-backslash them: \\{, or enclose your maths in some sort of "protection" scheme: <p>..</p> tags are best, if the paragraph doesn't have any other markup (such as displayed maths)) – Andrew Stacey Apr 19 2010 at 7:10