Continuous, bijective - yet not a homeomorphism

Question

I'm going through the earlier chapters in books and making sure I can do everything (and addressed many short-comings, like compactness) but I've come across something I can't do.

In "Introduction to Topological Manifolds" - by John Lee "Exercise 2.28" there is a question which I am stuck on, and confused about the wording.

It goes as follows:

Let $X$ be the half open interval $[0,1)\subset\mathbb{R}$ and let $\mathbb{S}^1$ be the unit circle in $\mathbb{C}$ - both with their Euclidian metric topologies.

Define a map $a:X\rightarrow\mathbb{S}^1$ by $a(s)=e^{2\pi js}$ (where $j=\sqrt{-1}$).

Show that $a$ is continuous and bijective but not a homeomorphism.

It includes a picture of [------) and a circle with -)[- at the "3-o'clock" position, with the [ facing vertically up. That's not really important here though.

I am (finally!) happy with the topology part. I know to implicitly consider the subspace topology. I can see it is clearly bijective (but I am not sure how to prove this, probably by saying "but there's only one solution on the interval $[0,2\pi)$ so I could do it)

But how do I show it is continuous and that its inverse isn't.

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What have I tried?

I recently did http://www.maths.kisogo.com/index.php?title=Continuity_definitions_are_equivalent - this shows that definitions of continuity are the same. I could scrape together some proof by cutting the circle into little open intervals and looking at the pre-image. Then show they generate the topology but this isn't a very general method.

The "weakest" (as in most specific) continuity definition (the metric space one where $d(a,b)=|a-b|$) is something I can do. I have no experience with trying to show the most general one. Continuity in terms of open sets, and I want to know what methods I can employ to show continuity.

Quick addendum: other examples are welcome I'd like to practice this (proving continuity between topological spaces) so any other examples are welcome.

Answer 1

A continuous real-valued function defined on a compact set is bounded. You can see that semi-closed interval is not compact by defining an unbounded continuous function. Then you can conclude that no map from a compact set to this set could be a homeomorphism.

Answer 2

Usually if you want to show continuity directly, as it seems like you want to do, you just check that the function is continuous on basis elements (see http://planetmath.org/testingforcontinuityviabasicopensets). Now basis elements for the topology on $\mathbb{R}$ are just open intervals, and for $\mathbb{R^2}$ they are open disks. The subspace topology then has a basis consisting of intersections of these intervals/disks.

Now for your specific case, the intersection of a disk and the circle is an arc of the circle. What is the inverse image of this arc? (Hint: points involving the north pole behave a little bit differently).

On the other hand, were this function to have a continuous inverse, what would be the inverse image of the open set $(0,1) \subset [0,1)$?