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Apologies for the vague title, but I can't describe the following question well.

The question is (as usual, from my book)

Let $f:X \to X'$ and $g:Y \to Y'$ be continuous. If $\mbox{cls} \varphi \in H^p(X';R)$ and $\mbox{cls} \theta \in H^q(Y';R)$, then $(f \times g)^*(\mbox{cls} \varphi \times \mbox{cls} \theta) = f^* \mbox{cls} \varphi \times g^* \mbox{cls} \theta.$

(here $R$ is a commutative ring). So the way I read the question, the multiplication is the cup product. I will first introduce the notation used in the book - I am not sure if it is standard.

If $\varphi \in S^n(X,G)$ and $c \in S_n(X)$, we write $$(c,\varphi)=\varphi(c) \in G.$$

Let $X$ be space, and let $R$ be a commutative ring. If $\varphi \in S^n(X;R)$ and $\theta \in S^m(X;R)$, define the cup product $\varphi \cup \theta \in S^{n+m}(X,R)$ by $$(\sigma,\varphi \cup \theta) = (\sigma \lambda_n, \varphi)(\sigma \mu_m,\theta)$$ for every $(n+m)$-simplex $\sigma$ in $X$, where the right side is the product of two elements in the ring $R$. ($\mu$ and $\lambda$ are the back face and front face functions respectively)

So I started looking at $f^* \mbox{cls} \varphi \times g^* \mbox{cls} \theta$. We have that $$ \begin{align} f^* \mbox{cls} \varphi \times g^* \mbox{cls} \theta &=\mbox{cls}(f^\# \varphi)\mbox{cls}(g^\# \theta) \\ &=\mbox{cls}([f^\# \varphi] \smile [g^\# \theta]) \\ &=\mbox{cls}\left((\sigma \lambda_p, f^\# \varphi)(\sigma \mu_q,g^\#\theta)\right) \\ &=\mbox{cls}\left((f \sigma \lambda_p,\varphi)(g \sigma \mu_q,\theta)\right) \end{align} $$ The other side is just $$ \begin{align} (f \times g)^*(\mbox{cls} \varphi \times \mbox{cls}\theta) &=(f \times g)^*\mbox{cls}(\varphi \smile \theta) \\ &=(f \times g)^*\mbox{cls}[(\sigma \lambda_p, \varphi)(\sigma \mu_q, \theta)] \end{align} $$

I am unsure how to proceed - specifically with the $(f*g)^*$ term. Now I know that $f:X \to Y$ yields a ring homormophism $f^*:H^n(Y;R) \to H^n(X;R)$ given by $f^* \mbox{cls} \varphi = \mbox{cls} f^\# \varphi$

Hopefully all my notation is here is enough to understand the question - if not, let me know and I will edit and update

Edit: As pointed out in the comments below, the above is wrong and the question is really about the naturality of the cross product. I guess my confusion with the $(f \times g)^*$ term still stands. Specifically I know that if $f:X \to X'$ then we have an induced chain map $f_\#:S_*(X) \to S_*(X')$.

We can then apply the functor $\mbox{Hom}(\quad,G)$ to show that $f^\#$ induces a homomorphism $$f^*:H^n(X';G) \to H^n(X;G)$$ given by $\mbox{cls} \varphi \mapsto \mbox{cls}(\varphi f_\#)$. But I don't know what the map $(f*g)^*$ is?. Obviously is is a map $$(f*g)^*:H_n(X' \times Y') \to H^n(X \times Y)$$, but where does it take the equivalence classes?

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    What do you mean by "cls"? In any case, this looks more like the cross product than the cup product to me.2011-05-10
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    @Aaron - "cls" just means (equivalence) class. Maybe it is the cross product - the previous section has both cross and cup product in it. But, notation wise the book usually uses $\otimes$ for cross product2011-05-10
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    I am pretty sure that this question is *not* about cup product, and is rather about cross product, as @Aaron Mazel-Gee suggests. Cup prodcut takes two cohomology classes on the same space, and produces a third class in the cohomology of this space. Cross product takes a class on each of $X$ and $Y$, and produces a class on $X \times Y$. Your question is evidently about the latter situation. More precisely, it is asking you to verify that the formation of the cross product is functorial with respect to the pair of spaces $X,Y$. If you figure out how to describe the cross product concretely,2011-05-10
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    which may involve working through Eilenberg--Zilber, the proof should be pretty straightforward. (I don't know exactly how things are developed in the text you're reading, but I am guessing that you are defining cohomology by singular chains, and using Eilenberg-Zilber to relate singular chains on a product to singular chains on each factor.)2011-05-10
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    @Matt - opps, yes you (and Aaron) are obviously correct! I'll have a go at this question now2011-05-10

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