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Hatcher defines the cohomology ring $H^*(X;R)$ for a space $X$ and commutative ring $R$ to be the direct sum of the cohomology groups $H^n(X;R)$, and by his definition of a graded ring, elements of $H^n(X;R)$ will have dimension $n$.

In his discussion of the cross product $H^*(X;R)\otimes_R H^*(Y;R) \rightarrow H^*(X \times Y; R); a \otimes b \mapsto a \times b$ where $\otimes_R$ denotes the tensor product as $R$-modules.

He says that this becomes a ring homomorphism if we define the multiplication in a tensor product of graded rings by $(a \otimes b)(c \otimes d)=(-1)^{|b||c|}ac \otimes bd$ where $|x|$ denotes the dimension of $x$.

My problem here is that $a$ and $b$ are general elements of their respective cohomology rings, so are finite sum $a=\sum{\alpha_{n_i}}$ where each $\alpha_{n_i}$ is in some cohomology ring $H^{n_i}(X;R)$ and similarly for $b=\sum{\beta_{n_j}}$ For example, if $a=\alpha_1 +\alpha_2$ where $\alpha_i \in H^i(X;R)$, then what is the dimension of $a$ supposed to be?

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You are right, a element $x \in H^*(X,R)$ does not have a fixed degree. But since multiplication should be bilinear, it's enough to define the multiplication for "pure" elements on the form $a \otimes b$ where $a,b$ are of fixed degree.

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  • $\begingroup$ Just to add: sometimes you see "the definition extends by linearity." This is code for "we are using the universal property of tensor products given this bilinear map defined on the basis vectors." $\endgroup$ – Kyle Miller Jun 25 '17 at 1:22
  • $\begingroup$ So how exactly does it extend from pure elements? $\endgroup$ – TuoTuo Jun 25 '17 at 2:08
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    $\begingroup$ @TuoTuo : Write $a = \sum a_i$ where $a_i$ are of pure degree and similary for other terms. We obtain $$( (\sum a_i) \otimes (\sum b_j) )((\sum c_k) \otimes (\sum d_l)) = \sum_i (a_i \otimes(\sum b_j) \otimes ((\sum c_k) \otimes (\sum d_l)) = \dots = \sum_{i,j,k,l} (-1))^{jk} a_ic_k \otimes b_j d_l $$ and the last expression contains only products of pure elements. $\endgroup$ – user171326 Jun 25 '17 at 6:58

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