I am looking into the definition of the linking number. I've considered these two definitions. Consider a link $L$ with components $K_1$ and $K_2$, and respectively their embeddings $\gamma_1$ and $\gamma_2$.
Firstly, $Lk^{(1)}$ defined by counting the positive and negative intersections of a diagram of the link $L$ with +1 and -1 respectively, ultimately dividing this number by 2.
Secondly, one can define the linking number of $L$, by $Lk^{(2)}$, by taking the degree of the map
$\Psi: S^1\times S^1 \to S^2: (s,t)\mapsto \frac{\gamma_1(s) - \gamma_2(t)}{|\gamma_1(s) - \gamma_2(t)|}$
We now know the linking number is symmetric w.r.t the component knots. However $Lk^{(2)}(K_2,K_1) = deg(\mu\circ\Psi)$. with $\mu:S^2\to S^2$ that maps points to their antipodal points. However the degree of $\mu$ is -1, and thus
$Lk^{(2)}(K_2,K_1) = deg(\mu\circ\Psi) = -deg(\Psi) = -Lk^{(2)}(K_1,K_2) $.
And thus $Lk^{(2)}$ is not symmetric. Where did I go wrong?
I would think that $Lk^{(2)}\ne deg(\mu\circ \Psi)$, since this refers to
$\Psi:S^1\to S^1: (s,t)\mapsto \frac{\gamma_2(t)-\gamma_1(s)}{||\gamma_2(t)-\gamma_1(s)||}$.
However $Lk^{(2)}$ refers to $\Psi:S^1\to S^1: (t,s)\mapsto \frac{\gamma_2(t)-\gamma_1(s)}{||\gamma_2(t)-\gamma_1(s)||}$
Here one sees that also the variables are switched and this adds an extra factor $-1$ to your degree. This should solve your sign problem.
Could someone confirm my idea is correct (comment or otherwise)? Since I am relatively new to these orientations and degrees and may have missed something.