# diagonal image of a primitive homology class

Let $X$ be a topological space. A class $a\in H_n(X;\mathbb Z)$ is said to be primitive if $a\not = m b$ for every integer $m>1$ and $b \in H_n(X;\mathbb Z)$. Let $$\Delta_*:H_n(X;\mathbb Z)\to H_n(X\times X;\mathbb Z)$$ be the induced map in homology of the diagonal map $\Delta(x)=(x,x)$. I want to show that if $a\in H_n(X;\mathbb Z)$ is primitive then $\Delta_*(a)=a\otimes 1+1\otimes a$

I know that in a field $F$ we can use a simplified Kunneth formula in the following way: $$H_n(X\times X;F)=\sum_{i+j=n}{H_i(X;F)\otimes H_j(X;F)}=H_n(X;F)\otimes H_0(X;F)\oplus H_0(X;F)\otimes H_n(X;F)\oplus \sum_{i+j=n,i\not =0,i\not =n}{H_i(X;F)\otimes H_j(X;F)}$$ But i can't go further!

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The You have got the Kunneth formula wrong. In general I don't believe that $\textrm{Tor}_1^{\Bbb{Z}} (H_p(X),H_q(X))$ vanishes. – user38268 Oct 25 '12 at 11:14
you are right, i will take coefficients in a field $F$. – palio Oct 25 '12 at 12:09
Is this true integrally? Isn't the diagonal on $H_*(\mathbb{C}P^\infty; \mathbb{Z}) = \mathbb{Z}\{x_i, i\geq 0\}$ given by $\Delta(x_n) = \sum_{i+j = n} x_i\otimes x_j$? – ruediger Feb 22 '13 at 13:24