Why is the cohomology ring of a H-space a Hopf algebra? Let $X$ be a finite-type H-space. This means $H^*(X)$ is finitely generated, and there exists a 'multiplication' $\mu:X \times X \to X$ and an 'identity' $e\in X$ such that the maps $x \mapsto \mu(x,e)$ and $x \mapsto \mu(e,x)$ are homotopic to identity map (Note that 'inverses' and 'associativity' are not required). I have heard that the cohomology ring $H^*(X)$ then has the structure of a Hopf algebra. However, I do not understand how to prove this.
To my understanding, a Hopf algebra $A$ is an algebra with a co-multiplication $\Delta: A \to A \otimes A$ and an antipode map $S:A \to A$ such that $\mu((S\times Id)(\Delta(x)))=\mu((Id\times S)(\Delta(x)))=1$. In our case, the cup product $H^*(x) \otimes H^*(X) \to H^*(X)$ can act as the multiplication, while the induced homomorphism $\mu^*:H^*(x) \to H^*(X) \otimes H^*(X)$ can act as the co-multiplication (Kunneth formula is used here). However, what will play the role of the antipode map $S$? A first instinct would be the homomorphism induced by the inverse, but H-spaces may not have inverses!
I am further confused by the fact that, upon searching the Internet, many sources prove the statement 'cohomology ring of a H-space is a Hopf algebra' by adapting a different definition of Hopf algebra than the one I have said above. This alternative definition is: a graded algebra $A$ with a co-multiplication $\Delta: A \to A \otimes A$ such that $\Delta(x)=1 \otimes x + x \otimes 1 + \Sigma x'_i \otimes x''_i$, where $0<|x'_i|,|x''_i|<|x|$. It is quite easy to show that $H^*(X)$ is a Hopf algebra of this definition, simply by following diagrams of induced homomorphisms, but it is unclear to me that these two definitions of Hopf algebra are equivalent (even if we require a grading in the first definition). 
Furthermore, if these two definitions are indeed equivalent, then it seems that the antipode map $S$ should be obtainable from the multiplication and co-multiplication in a natural way, which translates to a method of expressing it in terms of some map on the H-space $X$. However, I cannot see how any map would fit the role of $S$ aside from an inverse.
Any remarks or references will be appreciated!
 A: First note that if $X$ is an associative  H-complex then it has a unique inverse. If it is not associative then it has has unique left- and right-inverses and there are conditions under which these are equal. A reference is "The inverses of an H-space" by Arkowitz, Oshima, Strom.
In the general case an antipode may be defined degrewise since $X$ is of finite type.
A more general statement as pointed out by Qiaochu Yuan is contained in section 21.3 of May and Ponto's "More Concise Algebraic Topology".
I wouldn't worry too much about the slight differences in the definitions. There have been different general definitions of both Hopf algebras and H-spaces adopted across the literature. Algebraic topologists in particular are used to working with CW complexes and may brush over the finer points of the full theory to make their work more accessible, since they deal mainly with such a nice class of Hopf algebras. 
Certainly in the case of a finite (type), positively graded Hopf algebra the two definitions you give coincide. These are certainly the class studied by algebraic topologists. I believe the second definition has some difficulties without the assumption of boundedness.
For you last question note that the comultiplication on $X$ is not topological. It comes from the diagonal $\Delta:X\rightarrow X\times X$ followed by the (inverse of the) Kunneth isomorphism. Now the Kunneth isomorphim may be defined topologically by a map $H^*(X)\otimes H^*(X)\xrightarrow{\cong} H^*(X\times X)$ obtained by smashing together maps into Eilenberg Mac-Lane spaces. The inverse is defined in purely algebraic terms and I'm not sure how it would be used to produce a topological H-inverse for $X$ in the general case.
