Let our spaces have the homotopy type of a CW complex, and let $E^*,F^*$ be two cohomology theories. A (degree $n$, stable) cohomology operation is a map $\Phi: E^q(X) \to F^{q+n}(X)$ for each $q$, such that $\Phi \circ \Sigma= \Sigma \circ \Phi$ where $\Sigma$ is the suspension isomorphism.
In A Concise Course on Algebraic Topology, p. 184:
In general, cohomology operations are only natural transformations of set-valued functors. However, stable operations are necessarily homomorphisms of cohomology groups…
How does one check this?
Let $f,g \in E^q(X)$. We want to show that $\Phi(fg) = \Phi(f)\Phi(g)$.
My only idea is to use that $E$ and $F$ are representable by $\Omega$-spectra, so $f,g \in [X,E_q]$. Q1: But then what is $fg$, concretely?
Since we have an $\Omega$-spectrum, we have a weak homotopy equivalence $E_q \to \Omega E_{q+1}$, hence $f,g$ are homotopy classes of maps $[X, \Omega E_{q+1}]$, which is $\cong [\Sigma X, E_{q+1}] = E^{q+1}(X)$. I am implicitly composing with the weak homotopy equivalence. So we know that
$$\Phi_q(fg) = \Sigma^{-1}\Phi_{q+1} \iota_q(fg),$$
where $\iota_q$ is the weak equivalence $E_q \to \Omega E_{q+1}$. Then I want the RHS to be $\Phi_q(f)\Phi_q(g)$. But still, I need to know something about $fg$, which I don't!
Best Answer
For any space $Y$ there is an isomorphism $$\iota:H^*(Y\vee Y)\xrightarrow\cong H^*Y\oplus H^*Y,\qquad \gamma\mapsto (i_1^*\gamma,i_2^*\gamma),$$ where $$Y\xrightarrow{i_1}Y\vee Y\xleftarrow{i_2}Y$$ are the inclusions into each factor. The inverse is given by $$\kappa:H^*Y\oplus H^*Y\xrightarrow\cong H^*(Y\vee Y),\qquad (\alpha,\beta)\mapsto q_1^*\alpha+q_2^*\beta$$ where $$Y\xleftarrow{q_1}Y\vee Y\xrightarrow{q_2}Y$$ are the maps collapsing opposite factors to a point.
Below we'll need to know that these isomorphisms are compatible with the action of any cohomology operation $\Phi$. This is easy to check for $\iota$, for which naturality provides $$\iota(\Phi_{Y\vee Y}(\gamma))=(i_1^*\Phi_{Y\vee Y}(\gamma),i_2^*\Phi_{Y\vee Y}(\gamma))=(\Phi_Y(i_1^*\gamma),\Phi_Y(i_2^*\gamma))=\Phi_Y\oplus\Phi_Y(\iota(\gamma))$$ for any $\gamma\in H^*(Y\vee Y)$. For the inverse compute $$\Phi_{Y\vee Y}\circ\kappa=\kappa\circ\iota\circ(\Phi_{Y\vee Y}\circ \kappa)=\kappa\circ(\Phi_{Y\vee Y}\oplus\Phi)\circ(\iota\circ\kappa)=\kappa\circ(\Phi_Y\oplus\Phi_Y).$$
Now, specialise to the case $Y=\Sigma X$ for some space $X$. Recall that the suspension is equipped with a comultiplication $$c:\Sigma X\rightarrow \Sigma X\vee\Sigma X$$ satisfying $$q_1\circ c\simeq id_{\Sigma X}\simeq q_2\circ c.$$ Consequently, if $\alpha,\beta\in H^*(\Sigma X)$, then $$c^*(\kappa(\alpha,\beta))=\alpha+\beta.$$
Finally, suppose that $\Phi$ is a stable cohomology operation and $\alpha,\beta\in H^*X$. For notational reasons write $\alpha'=\sigma(\alpha),\beta'=\sigma(\beta)\in H^{*+1}(\Sigma X)$ for the suspensions of $\alpha,\beta$. Then $$\Phi_X(\alpha+\beta)=\Phi_{\Sigma X}(\alpha'+\beta')=\Phi_{\Sigma X}(c^*\kappa(\alpha,\beta))=c^*(\Phi_{\Sigma X\vee\Sigma X}(\kappa(\alpha',\beta')))=c^*\kappa(\Phi_{\Sigma X}\alpha',\Phi_{\Sigma X}\beta')=\Phi_{\Sigma X}\alpha'+\Phi_{\Sigma X}\beta'=\Phi_X\alpha+\Phi_X\beta$$ which is what was to be shown.