Consider the map $S^m\times S^n \rightarrow S^n\times S^m$ which takes $(x,y)$ to $(y,x)$. This induces a map on the smash product $S^m\wedge S^n=S^{m+n}$. I am at a loss thinking what the degree of this map should be. I suspect it should be $(-1)^{mn}$ since it is a homeomorphism and should be a power of $-1$ and I am kind of permuting the coordinates and that is the sign of the desired permutation but don't know how to prove it.
Any help will be highly appreciated. Thanks.
Concretely, the natural quotient map $S^m\times S^n \to S^m\wedge S^n \cong S^{n+m}$ is given by:$$ ((x_0,x_1,\cdots,x_m),(y_0,y_1,\cdots,y_n))\mapsto \left(\frac{x_1}{1-x_0},\cdots,\frac{x_m}{1-x_0},\frac{y_1}{1-y_0},\cdots,\frac{y_n}{1-y_0}\right) $$
On the left hand side here, we regard elements of $S^m$ (resp. $S^n$), as $m+1$- tuples (resp. $(n+1)$-tuples), which square sum to $1$. On the the right hand side we identify $S^{n+m}$ with $\mathbb{R}^{n+m}\cup \{\infty\}$.
Thus your map $S^{n+m}\to S^{n+m}$ is induced by the linear map $\mathbb{R}^{n+m}\to \mathbb{R}^{n+m}$ given by the matrix $$\left(\begin{array}{cc} 0 &I_n\\ I_m & 0 \end{array}\right)$$
Clearly this has determinant $(-1)^{(n+m+1)n}=(-1)^{nm}$.