Let $M^m$ and $N^n$ be compact, oriented smooth manifolds without boundary. Then what is the degree of the map
$ f: M\times N \to N \times M$
given by $f(x,y) = (y,x)$? I have the feeling it should be $(-1)^{mn}$ (would fit in with some proof I have to give), but the result I come up with is $1$.
My reasoning: If $w_1, \dots, w_m$ and $v_1, \dots, v_n$ are positively oriented ordered bases for $T_pM$ and $T_qN$, then $w_1, \dots, w_m, v_1, \dots, v_n$ is a positively oriented ordered basis on $T_{(p,q)}(M \times N)$ (where $T_{(p,q)}(M \times N) \simeq T_pM \oplus T_qN$).
Similarly, $v_1, \dots, v_n, w_1, \dots, w_m$ is a postiviely oriented basis for $T_{(q,p)}(N \times M)$. Now $df(p,q)(w_1, \dots, w_m, v_1, \dots, v_n) = (v_1, \dots, v_n, w_1, \dots, w_m)$, so $f$ is order preserving. And since it is a diffeomorphism, it's degree is $1$.
Now since the above would contradict another statement I have to prove, I must be making a mistake. I would be very glad if someone could help me out here. :-)
Best Regards,
S.L.