Generalized cohomology on the one point space

Question

I am reading Hatcher's algebraic topology for an assignment on generalized cohomology theories, and in section 4.E p. 447 he says the following

The wedge axiom implies that $h(\textit{point})$ is trivial. To see this, just use the fact that for any $X$ we have $X \vee \textit{point} = X$ , so the map $h(X)\times h(\textit{point})\to h(X)$ induced by inclusion of the first summand is a bijection, but this map is the projection $(a, b)\mapsto a$, hence $h(\textit{point})$ must have only one element.

I do not understand why this map must be the projection, since the axioms do not specify how to build the induced map in cohomology for a map of CW complexes.

Thanks you in advance

Answer 1

It is part of the wedge axiom: $X\vee Y$ is the coproduct in pointed spaces, and therefore comes with distinguished maps ("inclusions") from $X$ and $Y$, giving a coproduct diagram $X\rightarrow X\vee Y \leftarrow Y$. Applying $h$ gives a diagram $$h(X)\leftarrow h(X\vee Y)\rightarrow h(Y).$$ Now $h(X)\times h(Y)$ is the product in the category where $h$ takes values, and therefore comes with distinguished maps ("projections") to each factor, giving a product diagram $$h(X)\leftarrow h(X)\times h(Y)\rightarrow h(Y).$$ The wedge axiom says that not only is $h(X\vee Y)$ isomorphic to $h(X)\times h(Y)$, but it is isomorphic in a way that is compatible with two displayed diagrams, i.e. under that isomorphism the two legs of the top diagram become the two legs of the bottom diagram.

Answer 2

A prettier way of getting at this is to say that, if $S = h(point)$, then the isomorphism of the wedge axiom applied to $point \vee point$ says that the diagonal map $S \rightarrow S \times S$ will be a bijection. But this implies $S$ has only one element.