Select Page

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.