Tor term in the homology of a product of real projective spaces

The pairs of nonnegative indices with sum three are $(0,3),(1,2),(2,1),(3,0)$. Using [F1], their tensor groups are respectively $\mathbb{Z},0,0,\mathbb{Z}$: tensor with zero is zero, and $\mathbb{Z}\otimes \mathbb{Z}\cong \mathbb{Z}$ by multiplication with inverse $a\mapsto a\otimes 1$. Thus the tensor diagonal is $K={\mathbb{Z}}^2$, generated by the top class of each factor crossed with a point in the other.

The Tor indices have sum two: $(0,2),(1,1),(2,0)$. The outer pairs have a zero input and hence zero Tor, computed with the zero resolution in that variable. For the middle pair use the exact free resolution $0\to \mathbb{Z}\stackrel{2}{\to }\mathbb{Z}\to \mathbb{Z}\mathrm{/}2\to 0$. Tensoring with $\mathbb{Z}\mathrm{/}2$ makes its only differential multiplication by two, hence zero. Its degree-one homology is the kernel of that zero map, all of $\mathbb{Z}\mathrm{/}2$, with no incoming degree-two boundary. Therefore $Q={\mathrm{Tor}}_1(\mathbb{Z}\mathrm{/}2,\mathbb{Z}\mathrm{/}2)=\mathbb{Z}\mathrm{/}2$.

Substitution in [F2] gives $0\to {\mathbb{Z}}^2\stackrel{\times }{\to }{H}_3(X\times Y;\mathbb{Z})\stackrel{\gamma }{\to }\mathbb{Z}\mathrm{/}2\to 0$. Let $j$ be a section supplied there. Then $t=j(\bar{1})$ is nonzero since $\gamma t=\bar{1}$, and $2t=j(0)=0$. Every class $w$ has a unique expression $\alpha (k)+j(q)$: take $q=\gamma w$, then $w-j(q)$ belongs to the image of the injective cross product $\alpha$. Uniqueness follows by applying $\gamma$ and then injectivity of $\alpha$. This proves the asserted direct sum and exhibits its nonzero order-two class. Its choice depends on the section; the nonzero Tor quotient does not.

The witness $t$ cannot belong to the tensor image, since that image is the kernel of $\gamma$ while $\gamma t=\bar{1}\mathrm{\ne }0$. The two free generators cannot cancel this correction. The zero element of the Tor group lifts to zero under $j$, and its sole nonzero element has order exactly two. The dimensions are fixed at three; no claim is made by substituting ${\mathbb{R}\mathbb{P}}^0$ or an empty factor. AC is inherited from the Kunneth sequence and its splitting, and from its balanced-Tor convention; these finite resolution calculations introduce no additional choices.