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?
 A: 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: 


*

*$h$ is 1-1 

*$h$ is onto

*$h$ is continuous

*$h^{-1}$ 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.
A: 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: 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.
