Symmetry of the linking number (Problem 13 of Milnor's Topology from the differentiable viewpoint).
Let $M^n,N^m\subset \mathbb{R}^{m+n+1}$ be compact, oriented differential manifolds without boundary of dimension $n$ and $m$.
Consider
$\phi:M^n \times N^m\rightarrow S^{n+m}:(p,q)\mapsto \frac{q-p}{||q-p||}$
and define $l(M,N)=$deg$(\phi)$.
I would like to prove $l(M,N)=(-1)^{(m+1)(n+1)}l(N,M)$.
So far I did the following. Define 
$\phi_1:M^n \times N^m\rightarrow S^{n+m}:(p,q)\mapsto \frac{q-p}{||q-p||}$
and
$\phi_2:N^m \times M^n\rightarrow S^{n+m}:(q,p)\mapsto -\frac{q-p}{||q-p||}$.
We want deg$(\phi_1)=$deg$(\phi_2)$. Let $z$ be a regular value of $\phi_1$. If I am correct this means that $-z$ is a regular value of $\phi_2$. Moreover $\phi_1^{-1}(z)$ and $\phi_2^{-1}(-z)$ are in one to one correspondence via $(p,q)\mapsto (q,p)$. Therefore we are left to show for each such $(p,q)$:
sgn$(\phi_1,(p,q)) = (-1)^{(m+1)(n+1)}$sgn$(\phi_2,(q,p))$
where sgn$(\phi_1,(p,q))$ is +1 if the derivative map $(d{\phi_1})_{(p,q)}$ preserves orientation and -1 otherwise.
I am stuck proving this last equality. It looks like the antipodal map but theres is some change of coordinates going as well. I think I am messing up working with different orientations. Does anyone know how to proceed? Thank you for your help!
 A: As Milnor suggests, let the linking map be
$$\lambda:M^m\times N^n\to S^k,$$
$$\lambda(p,q)=\dfrac{p-q}{||p-q||},$$
with $m+n=k$, where $M,N\subset\mathbb{R}^{k+1}$ are disjoints, oriented, boundaryless manifolds. Define
$$\lambda':N^n\times M^m\to S^k,$$ 
$$\lambda'(q,p)=\dfrac{q-p}{||q-p||}=-\dfrac{p-q}{||p-q||}$$
the linking map of interest. Let $\phi:S^k\to S^k$ be the antipodal map. We see that we have $\lambda(p,q)=(\phi\circ\lambda')(q,p)$ for all $p\in M$ and $q\in N$. Taking into account the reversal map given by 
$$ \psi:M^m\times N^n\to N^n\times M^m$$
which maps $(p,q)$ to $(q,p)$ and has degree $(-1)^{mn}$, we can proceed. We have $\lambda=\phi\circ\lambda'\circ\psi$.
Thus, as problem 1 of Milnor suggests, 
$$\mathrm{deg}(\lambda)=\mathrm{deg}(\phi)\mathrm{deg}(\lambda')\mathrm{deg}(\psi).$$
Hopf theorem of section 7 and the end of section 5 of Milnor can help you now. We know that for $k$ odd, the antipodal map is homotopic to the identity but for $k$ even, it isn't. So, it is easily shown that
$$\mathrm{deg}(\phi)=(-1)^{k+1}.$$
It then follows that
$$\mathrm{deg}(\lambda)=(-1)^{n+m+1+mn}\mathrm{deg}(\lambda')=(-1)^{(m+1)(n+1)}\mathrm{deg}(\lambda')$$
which is what we wanted to show.

EDIT: I think I need to further explain what is up with this reversal map. 
Problem 8 in milnor shows that for two smooth manifolds, we have
$$T_{(x,y)}(M\times N)=T_x M\times T_y N.$$
With this in mind, the differential of the reversal map can be formulated as
$$(v_1,\dots, v_m,w_1,\dots, w_n)\mapsto (w_1,\dots w_n,v_1,\dots, v_m).$$
You can show that the determinant of this linear application is $(-1)^{mn}$ by counting permutations. I hope this helps.
