What is the homeomorphism type of the surface given by the polygonal presentation $aaa$? More precisely, I am interested in the mapping cone of the map $S^1 \to S^1,$ $z \mapsto z^3.$ It seems like it should yield a "surface" with the following polygonal presentation:

What is this space, exactly?
 A: 
What is this space, exactly?

The cone $X = Cf$ is the quotient space $Mf/X\times\{0\}$ obtained from pinching the top of the mapping cylinder to a point. Since $Mf$ is the cylinder $S^1 \times [0, 1]$ with the bottom pasted to $S^1$ by $f$, $Cf$ is the $2$-dimensional cell complex $D^2 \cup_f S^1$.
This is formally called the $3$-fold dunce cap, and the fundamental polygon you depicted is indeed correct. It's a generalization of the projective space $\Bbb RP^2$ in the sense that mapping cone of the two fold covering map $f : S^1 \to S^1$ is $\Bbb RP^2$. 
$\pi_1(X) = \langle a | a^3 = 1\rangle \cong \Bbb Z/3$ as the loops given by the words $1, a, a^2$ at the boundary of the cap are all distinct, and $a^3$ can be homotoped to a point by sliding through the interior of the disk.
However, as noted by Olivier Bégassat in the comments, $X$ is not a $2$-manifold : any open neighborhood of a point lying on the boundary of the triangle $aaa$ is homeomorphic to three half-disks pasted by the diameter, which is not homeomorphic to $\Bbb R^2$.
A: Is the following computation of $H_*(X)$ correct?
Since $X$ is a $2$-dimensional cell complex, $H_n(X) = 0$ for $n \geq 3.$
Taking $A = $ the boundary circle of the fundamental polygon, the long exact sequence of a pair yields $$0 = H_2(A) \to H_2(X) \to H_2(X,A) \to H_1(A) \cong \mathbb{Z};$$ I believe $(X,A)$ is a good pair, so that $H_2(X,A) \cong H_2(X/A) \cong H_2(S^2) \cong \mathbb{Z}.$ I believe that the fundamental polygon, viewed as a relative 2-cycle, generates $H_2(X,A),$ and is mapped to the generator of  $H_1(A)$ under the boundary map. Hence the rightmost arrow is injective, which implies that $H_2(X) = 0.$
Finally, $H_1(X) = \pi_1(X) = \mathbb{Z}/3$ and $H_0(X) = \mathbb{Z}.$
