Question:
True/False?
There exists a continuous surjective function from \(S^1 \) onto \(\mathbb{R}\).
Hint:
Search for topological invariants.
Discussion:
We know that continuous image of a compact set is compact. \(S^1\) is a subset of \(\mathbb{R}^2\), and in \(\mathbb{R}^2\) a set is compact if and only if it is closed and bounded.
By definition, every element of \(S^1\) has unit modulus, so it is bounded.
Let’s say \(z_n\to z\) as \(n\to \infty \). Where {\(z_n\)} is a sequence in \(S^1\). Since modulus is a continuous function, \(|z_n| \to |z| \), the sequence {\(|z_n|\)} is simply the constant sequence \(1,1,1,… \) hence \(|z|=1\).
What does above discussion mean? Well it means that if \(z\) is a limit point (or even a point of closure) of \(S^1\) then \(z\in S^1\). Therefore, \(S^1\) is closed.
The immediate consequence is that the given statement is False. Because, \(\mathbb{R}\) is not compact. \(S^1\) is compact, and continuous image of a compact set has to be compact.
