Consider the map $f:S^3 \times S^3 \to S^3 \times S^3$ given by $(z_0,z_1)\mapsto (z_1, -z_0)$. Write $f$ as composition of $g \circ h $, where $h$ is $(z_0,z_1)\mapsto (z_0, -z_1)$ and $g$ is $(z_0,z_1)\mapsto (z_1, z_0)$. Then the corresponding map induced by $h$ on $3^{rd}$ cohomology group with $\Bbb Z$-coefficient is given by $A=\pmatrix{1&0\\0&1}$(as the degree of the antipodal map is $(-1)^{n+1}=1$ in this case) and that of $g$ is given by $B=\pmatrix{0&1\\1&0}$. So the induced map by $f$ on $3^{rd}$ cohomology group is $AB$ i.e. $\pmatrix{0&1\\1&0}$. But this must be wrong as then the Lefschetz number $L(f)= 1+ (-1)^3 \cdot0+ (-1)^6\cdot1=2\neq 0$. But $f$ is a free action. So contradiction. Please help me to find my mistake.
I am learning Algebraic Topology by myself. Sorry, incase this a trivial question.
Thank you so much for your help.
The mistake is that the degree of the map is $-1$, not $1$. If $\alpha$ is the generator of the top cohomology of the first factor and $\beta$ that of the second, then note that $\pi_1^*\alpha\cup\pi_2^*\beta = (-1)^{3\cdot 3}\pi_2^*\beta\cup\pi_1^*\alpha$.