distinction between homeomorphic topological spaces I am trying to distinguish between topological spaces, given the following image:



I think that space $A$ is homeomorphic to space $B$, the function just need to "compact" the semi-plane to a circle, and map the frontier points of the axis to the semi-circumference of the circle. On the other hand i see that between spaces $A$ and $C$ there is not the same number of frontier points, that discrepancy will make impossible to construct a bijective between them. But i am not sure about this arguments, any observation will be greatly appreciated.
 A: Regarding $A$ and $B$, you're on the right track, but still there is some hard work to do regarding writing down a formula for the homemorphism.
Regarding $A$ and $C$, your wording "...there is not the same number of frontier points..." is inappropriate, as said the remark of @CameronBuie. But there is a way to alter the wording to make a correct statement, namely: 

... there is not the same number of connected components of boundary points ...

Since I do not know whether you have the background to understand how to make this rigorous, I'll simply list a few concepts that you need to learn for that purpose: 


*

*Manifolds with an emphasis on manifolds with boundary, and specialized to manifolds of dimension 2 also known as surfaces. It's important to be aware that in the context of a manifold-with-boundary, the "boundary" is quite different from the "frontier" which you mentioned in your post.

*Connectivity and connected components of topological spaces.

A: The proof that $A$ and $C$ are not homeomorphic requires some knowledge. 
As Lee Mosher pointed out, you can do it by looking at the boundary components of the manifolds $A, C$.
Another approach is to consider the one-point compactifications $A^+$ of $A$ and $C^+$ of $C$. If $A,C$ would be homeomorphic, then so would be $A^+ , C^+$. But $A^+$ is homeomorphic to a closed disk which is contractible and $C^+$ retracts to a copy of the circle $S^1$ which is not contractible. This is impossible.
