Problem:
Show that homeomorphism is an equivalence relation on topological spaces. Now consider the capital letters of the alphabet $\mathsf{A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z},$ in a sans serif font. Each of these gives a graph in the plane. Sort these into homeomorphism classes. [The partition may depend on the font! In particular, $\mathsf{K}$ can be tricky.]
What I got so far:
$\mathsf{C\cong G\cong I \cong J \cong L \cong U \cong V \cong Z}$
They are not closed curves and can be curved and bent to one another.
$\mathsf{D\cong O}$
Closed curves that can easily be bent to one another.
$\mathsf{C \ncong O}$ since removing a point in $\mathsf{C}$ makes it disconnected while $\mathsf{O}$ stays connected.
$\mathsf{E\ncong O}$ for the same reason.
$\mathsf{E\cong F}$
The standing line on $\mathsf{E}$ can be shrunk and the bottom line rotating and shrinking to make the straight standing line for $\mathsf{F}$. And the same reverse process for $\mathsf{F}$, which are homeomorphisms.
$\mathsf{K\cong X}$, this is easy to see.
and $\mathsf{K \ncong H}$ since deleting a point in $K$ gives 4 components but not in $H$.
I don't see any congruent letters for $\mathsf{A}$ and $\mathsf{H,R}$, so I think each is alone.
$\mathsf{P\cong Q}$ is clear.
$\mathsf{E\cong F\cong T \cong Y}$ as well.
$\mathsf{M\cong N\cong W}$ is also easy to see.
Is this correct, I would appreciate any corrections.