In JP May's [Concise Course in Algebraic Topology](http://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf "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 $t$...



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


