2 colorability of link diagram

Question

Prove that the shadow of any link diagram can be checkerboarded (2 colored).

Is there an elementary proof for this result (not requiring advanced theory)

Answer 1

The shadow of an $n$-component link diagram can be obtained from the standard, zero-crossing diagram of the $n$-component unlink via a sequence of shadows of Reidemeister moves. This diagram of the $n$-component unlink is checkerboard colorable (color the insides of the components black and the unbounded face white).

One can show that a checkerboard coloring before a shadow of a Reidemeister move induces a checkerboard coloring after the shadow of a Reidemeister move. Therefore, any link diagram is checkerboard colorable.

Answer 2

An argument I rather like for this as follows. Replace each crossing with an arbitrary smoothing:

After these smoothings, the shadow diagram is a collection of non-intersecting simple closed curves, possibly nested. Such a situation is $2$-colorable: it follows from the Jordan curve theorem that, if you start in a region and count how many times you cross a curve to get to the outermost region, no matter the path you took it will always be the same modulo $2$. Color the regions according to the modulus. (I'm just saying a reason why it is you can start coloring regions without worry.)

Now, unsmooth the crossings.