Nonexistence of a continuous injection $f:S^2 \rightarrow \mathbb{R^2}$

Question

What is the "easiest" way to show that there is no continuous injection $f:S^2 \rightarrow \mathbb{R^2}$?

Sure the Borsuk-Ulam theorem implies that result, but this may be a "difficult" way.

Answer 1

This is a non-obvious result. By invariance of domain, if there were such an $f$, its image would be open in $\Bbb R^2$. But it would, of course, also be compact. Oops.

Answer 2

A broad-strokes proof outline:

Consider two points $a,b \in S^2$ and a continuous closed curve $C\in S^2$ such that all coninuous transformations mapping $a\to b$ contain a value on $C$. The existtance of such a triplet is shown by example: The North and South poles, and the equator, on a sphere.

Now consider any continuous injection of that triplet onto some co-domain $D \in \Bbb{R}^2$. (Since the injection is also surjective onto $D$ it is a bijection.) Since every continuous curve line from $a$ to $b$ intersects $C$, and continuous injections preserve that property, exactly one of $a',b'$ must lie in the interior of $C'$. Without loss of generality, say $a'$ is not in the interior of $C'$.

Then $C'$ cannot be continuously deformed to lie completely in an arbitrarily small $\epsilon$-ball containing $a'$ in its interior. And a continous injection would preserve that property. Yet $C$ can be continuously deformed to lie completely in an arbitrarily small $\epsilon$-ball containing $a$ in its interior. This contradictoin shows that we could not have the desired continuous injection.