# Fundamental polygon of surface with boundary

I was playing with some shapes recently and came across a surface $$M$$ with the following fundamental polygon: link

This has surface word $$a c b f_1 a^{-1} c^{-1} f_2 b^{-1} f_3,$$

where the $$f_i$$ are boundary edges that are not identified to any other. I'd like to know how to identify this surface. I know this might be orientable because every paired edge has an inverse. I also see that there are two boundary circles $$\mathbb{S}^1$$ formed by $$f_1 f_2$$, and $$f_3$$ (these connect back up to themselves). Under identification, I count 3 vertices and 5 edges ($$a, b, c, (f_1 f_2),$$ and $$f_3$$). Given there is 1 face, this should have $$\chi (M) = 3 - 5 + 1 = -1$$ I recall that we can cap the two boundary circles to form a surface $$M^*$$ without boundary, which has $$\chi (M^*) = \chi (M) + 2 = 1$$ This doesn't fit with the classification of compact surfaces however. If it's orientable, then $$\chi (M^*)$$ would be even. Further, wouldn't $$\chi (M^*) = 1$$ imply that $$M^* \cong \mathbb{P}^2$$, the projective plane?

I don't actually know what I'm doing here. I know there might be some argument using word operations, or cutting and pasting, but I don't know where to start. Moreover, I think a larger question I have is how to even deal with unpaired edges in fundamental polygons in the first place. I've searched around quite a bit, but everything I've found just deals with surfaces without boundary. The best I've found is that the classification theorem extends to compact surfaces with boundary in a way that states that these surfaces are homeomorphic to any of $$\mathbb{S}^2, n\mathbb{P}^2,$$ or $$n\mathbb{T}^2$$ with a finite number of disks removed.

Any pointers or resources on this would be greatly appreciated. If anyone is interested, the place this turns up is from a very convoluted maze level in the 1980 Atari game Adventure. I think it'd be fun to figure out what surface this lies on!

There are six edges, not five: there is not one edge $$(f_1f_2)$$, instead there are two edges $$f_1$$ and $$f_2$$ which meet at a vertex.
So $$\chi(M)=-2$$ and $$\chi(M^*)=0$$. The surface $$M^*$$ is therefore a torus, and $$M$$ itself is a torus with two holes punched out of it.