Here is an outline of the proof of the Borsuk-Ulam Theorem; more details can be found in Section 2.6 of Guillemin and Pollack's book Differential Topology.

As there, we will deal with smooth maps, and make use of standard results like Sard's theorem.

Remember that Borsuk-Ulam says that any odd map f from Sⁿ to itself has odd degree. Here f is called odd if it is equivariant with respect to the antipodal map: f(-s)=-f(s).

The proof is by induction on n. For n=1, lift the map f from S¹ to S¹ to a map h from R to R with h(x+1)=h(x)+deg(d). But if f is odd, we find that h(x+1/2)=h(x)+k/2 for some odd integer k, and it then follows that deg(f)=k is odd.

For the inductive step, let k be the degree of f, and let g be the restriction of f to the equator S^n-1. By Sard's theorem, we can find a point a in the image sphere which is a regular value for g and f, meaning that it is not in the image of g and it is achieved exactly k times by f. After a rotation, we can assume that a is exactly the north pole of the sphere.

Because f is odd, k can be computed as the number of preimages in the northern hemisphere of the north or south pole. Defining f⁺ as the restriction of f to the northern hemisphere, composed with projection to the equatorial plane, we find that k is the number of preimages of 0 under f⁺.

Now neither north nor south pole is in the image of g, so we can retract g onto the equatorial sphere and get an odd map from S^n-1 to itself. Applying the inductive hypothesis, this map has odd degree.

But g is the restriction of f⁺ to the boundary. By a standard lemma in differential topology its degree is then the number of preimages of the regular value 0. (If we restrict f⁺ to small spheres around each preimage, they have winding number one each.) Thus we see that k must be odd.