What are all topological spaces obtained by gluing the edges of a triangle?

I am currently learning about polygonal presentations of surfaces.

In the notation I'm using (following Lee's "Topological Manifolds"), $\langle a, b \ |\ aba^{-1}b^{-1}\rangle$ is a presentation of the torus $\mathbb{T}^2$, and $\langle a,b\ |\ abab \rangle$ is a presentation of the real projective plane $\mathbb{P}^2$. Both of these examples can be thought of as specifying labellings and orientations of the edges of a square, which in turn specify how to glue the edges together to obtain the respective surfaces.

As a fun exercise for myself, I'm trying to list all possible topological spaces (surfaces?) that can result from gluing together the edges of a triangle. I conjecture that the following five four presentations represent all possible such spaces (up to homeomorphism), and also conjecture that they fall into the given homeomorphism classes: $$\langle a \ | \ aaa\rangle \approx \text{?}$$ $$\langle a \ | \ aaa^{-1}\rangle \approx \text{?}$$ $$\langle a, b \ | \ aab\rangle \approx \text{?}$$ $$\langle a, b, c \ | \ abc\rangle \approx \mathbb{D}^2 \text{ (closed disk)}$$

Questions: Are these, in fact, all of them (up to homeomorphism), or are there some that I've missed? Are any two on this list homeomorphic (meaning I've double-counted)? And are there any common descriptions of the homeomorphism classes with question marks? (I realize that "common descriptions" is vague.)

EDIT: By "five" I of course meant "four." That is, $$\langle a, b \ | \ aa^{-1}b\rangle \approx \mathbb{D}^2,$$ which is geometrically clear upon drawing the picture.

Note: These are in fact polygonal presentations, and not group presentations. Because we are dealing with triangles (which are in some sense degenerate), we cannot always read off the fundamental group directly from the polygonal presentation as if it were a group presentation. The example in the "EDIT" illustrates this.

Solution 1:

$aab$ is a Möbius band. $aaa$ is not a surface, nor is $aaa^{-1}$, as with three things coming together you're not locally homeomorphic to a plane or half-plane.

EDIT: In the comments, I mention one can prove that $aab$ is a Möbius band by invoking the theorem on classification of surfaces (although I accidentally wrote characterization instead of classification). Another way to do it involves surgery. First note that $aab=aabc$. Then draw a line from the $aa$ corner to the $bc$ corner, and cut along that line to get two triangles, $adc$ and $abd^{-1}$. Now put the two triangles back together, but identifying the edges labeled $a$; you get $dcdb^{-1}$, which you recognize as a Möbius band.

Solution 2:

In regards to $aaa$ this has a somewhat common description as the pseudo projective plane of order $3$. In general the pseudo projective plane of order $n$ can be defined in a similar way by taking a regular $n$-gon with the presentation $a^n$. Also my topology professor referred to $aaa^{-1}$ as the dunce cap space.