# Homeomorphism definition

I was told by my professor that homeomorphisms are continuous maps with continuous inverse, but do those conditions also imply that the map is bijective?

• There are three properties at play: 1) The map is continuous. 2) The map has an inverse. 3) The inverse is also continuous. One of those properties in particular is both a necessary and sufficient condition for the map to be bijective. – pjs36 Feb 26 '16 at 5:23
• In a word, Yes. – BrianO Feb 26 '16 at 5:38
• Only bijective functions can exhibit true inverses. – IAmNoOne Apr 2 '18 at 6:34

It was inconvenient that your professor worded it this way. From the get-go, saying that a homeomorphism is continuous function with a continuous inverse assumes that we have some function $f$, and that its inverse, $f^{-1}$, exists. Off-the-bat, just cause we have a function $f$, it does not mean that it has an inverse.

In addition, the way your professor phrased it doesn't make it clear whether we are talking about a subset of the range, or the whole range itself: in a homeomorphism, we must have the whole range, which is something your professors phrasing neglected to capture.

If I were you, I would just forget about what your professor said. It is kind of circular. Just to make it clear. By definition, a homeomorphism is a function $h$, from a topological space $X$, to a topological space $Y$, such that the following hold:

1. $h$ is 1-1
2. $h$ is onto
3. $h$ is continuous
4. $h^{-1}$ is continuous

Note, we do not say $h^{-1}$ exists here: this is a consequence of $h$ being 1-1, as we can always create $h^{-1}$ in such a case. ALSO, we imply that the image of $h$ is all of its range: this is captured by saying that $h$ is a bijection.

• Suppose $f: X \rightarrow Y$ is continuous. Suppose there exists $g: Y \rightarrow X$ is continuous satisfying $g \circ f$ is the identity function on $X$ and $f \circ g$ is the identity function on $Y$, then is $f$ a homeomorphism? – user317935 Feb 26 '16 at 6:08
• @user317935 Think abut that one for a little longer: Are $X$, $Y$ topological spaces?!? – 9301293 Feb 26 '16 at 6:09

Yes, if a function is a homeomorphism, it is bijective. The converse is not true: consider the function that wraps $[0,1)$ around the unit circle (can you define this? Why is this interval half opened?) Here, one of the conditions for a homeomorphism is violated.

Similarly, consider the function that takes the step function to the x-axis. This is continuous, but its inverse is not. Homeomorphisms preserve properties that we care about, such as connectedness.

A function that has an inverse is injective, or one-to-one. Every one-to-one function is onto some set. So a continuous function with a continuous inverse is a homeomorphism onto some set even if there is some larger target set.

However, if $f:A\to B$ is continuous and has an inverse and its inverse is also continuous, that does not mean $f$ is a homeomorphism from $A$ to $B$. Rather, $f$ is a homeomorphism from $A$ to some subset of $B$, which may or may not be all of $B$.

Among the simplest instructive examples of a continuous bijection whose inverse is not continuous is $\theta\mapsto e^{i\theta}$ from $[0,2\pi)$ to the circle.

• Every function with an inverse is a bijection. Are you talking about a one-sided inverse? – Thomas Andrews Feb 26 '16 at 5:30
• @ThomasAndrews : I'm talking about an inverse whose domain is some subset of $B$, which need not be all of $B$. – Michael Hardy Feb 26 '16 at 5:32
• So, you are using words loosely in a way that will confus a reader just learning terms? Given a function from $A$ to $B$, an inverse is, amongst other thing, a function from $B$ to $A$. Using the term to mean something else is only going to confuse a beginner, either now or later. – Thomas Andrews Feb 26 '16 at 5:34
• It is used in set theory, in topology. It is used in all standard math courses that I have ever taken. – Thomas Andrews Feb 26 '16 at 5:39
• It seems to be common convention be that if someone writes "$f\colon A\to B$ has an inverse", then the domain of that inverse is implicitly $B$. It's at best misleading to say "$f\colon A\to B$ is invertible" if all one means is that $f$ is injective. – BrianO Feb 26 '16 at 5:42