Let $\bigtriangleup ^2$ denotes the standard 2-simplex in $R^3$. I am wondering what is the quotient space of this simplex if we identify all three edges to a single edge? I guess it would be 2-sphere since in the case of $\bigtriangleup^1$, the resulting quotient space is $S^1$.
But when computing the homology group, I found the kernel of boundary map $\partial_2$: $\bigtriangleup^2 \to \bigtriangleup ^1$ is trivial, in fact, the image is the identified edge. Thus, the second homology group $H_2$ is zero, which is not the result we already know for 2-sphere.
So I think, the problem is the remaining edge. If it can be shrink to a point, then everything goes through. But I am not quite sure. Hope someone could help. Thanks!
IF your vertices are $A,B,C$ and you identify (linearly, as directed edges) $AB$ with $BC$, and then $BC$ with $CA$, then you've certainly got something nonorientable, even after the second step, and non-manifold-like along the edge.
If you identify $AB$ with $AC$ linearly (as directed edges), the result is still (topologically) a disk; depending on how you then identify $BC$ with the resulting line-segment in the disk, I suppose you might get varying results.
Regardless, it seems safe to say that without more details, your question doesn't define a single object, so it's tough to say what that object is.