Prove/disprove that $X$ is a manifold

Question

Let $X$ be the 2-dimensional CW complex where 2-cells are glued as follows:

The goal is to prove/disprove that $X$ is a manifold. Here's my attempt to disprove this.
Suppose $X$ is a manifold. In particular, $X$ is a closed 2-dimensional manifold.
Since $H_2(X) \cong \mathbb Z$, it follows that $X$ is orientable. Therefore, by Poincare duality, $H^1(X) \cong H_1(X) \cong G$, where $G$ is the abelian group generated by five elements $a, b, c, d, e$ satisfying $a+b+c+d+e = 0$.
By Universal coefficient theorem, $H^1(X)$ must be a free abelian group, and so $G$ must be also free. Then, I attempted to show that $G$ has a non-zero torsion element which would yield a contradiction.

I would appreciate any hint or help with proving/disproving this statement.

Answer 1

Your space $X$ is indeed a 2-dimensional manifold.

There is a general theorem in 2-dimensional topology which goes like this:

If $X$ is the CW complex formed as the quotient of a finite collection of disjoint convex Euclidean polygons by identifying edges in pairs using a choice of edge pairing homeomorphisms [as your example is] then $X$ is a compact 2-manifold.

It's not too hard to see that $2$-cells of $X$ correspond one-to-one with the given polygons, and the $1$-cells correspond one-to-one with the given edge pairs.

The hardest part of the proof of this theorem is to identify the $0$-cells of the CW structure on $X$, and to verify that the image in $X$ of each vertex of the polygon collection has a 2-manifold neighborhood. The proof requires an understanding of "vertex cycles". The 0-cells of the CW structure will correspond one-to-one with the vertex cycles. And for each vertex in the quotient, a disc neighborhood of that vertex is obtained by gluing together "circular sector neighborhoods" of the elements of the associated vertex cycle.

In this example, the ten vertices form a single vertex cycle, which you can see as follows.

Start with the upper vertex of the left polygon, and label it by the unordered pair $\{e+,a-\}$ meaning that it is the terminal vertex of a certain $e$ edge and the initial vertex of an $a$ edge.

To walk to the next step of the vertex cycle, look for another vertex with label $a-$: that would be the upper vertex of the right polygon, labelled $\{a-,c+\}$ (it is the initial vertex of an $a$ edge and the terminal vertex of a $c$ edge).

Next, look for another vertex with label $c+$: that would be the lower left vertex of the left polygon, labelled $\{c+,d-\}$.

Continuing this process, one gets the vertex cycle \begin{align*} &\{e+,a-\}, \{a-,c+\}, \{c+,d-\}, \{d-,a+\}, \{a+,b-\},\\ &\{b-,d+\}, \{d+,e-\}, \{e-,b+\}, \{b+,c-\}, \{c-,e+\}, \\ &\{e+,a-\} \end{align*} which is a length $10$ vertex cycle, starting at and returning to the upper vertex of the left pentagon labelled $\{e+,a-\}$. So, the single vertex of the quotient has a 2-disc neighborhood formed by gluing together $10$ circular sector neighborhoods in a circular pattern, one such sector neighborhood for each of the $10$ vertices of the diagram.

Answer 2

I think it's much simpler than you're making it. Let's look at the "corners" and edges. From the right diagram, we see that the end of $a$ is the start of $d$; from the left, the end of $a$ is the start of $b$. From the right diagram, then end of $d$ is the start of $b$. Taken together, this means that the start and end of $d$ are in fact the same, and $d$ is a loop at a single location we could call $P$. A bit more fiddling like this means that all five vertices shown here are in fact a single vertex in the quotient (I think...I haven't actually checked all the details). Hold that thought.

In the quotient, all interior points of the two disks shown are clearly manifold-like (i.e., they have a locally-euclidean neighborhood). So the only question is whether $P$ has such a neighborhood. From a little bit of sketching, I strongly suspect it does not. But you could check that: remove the center 90% of each of the two pentagons. That gives you two strips which, joined along the edges as shown, should form a topological disk. So start drawing (or gluing up thin strips of paper) to see what's actually there.

(Presumably you could actually say that a neighborhood of the centers of the two pentagons is some set $A$ in the total space, while $X$ is the total space, and compute the homology $H_{*}(X, A)$ or something like that (or maybe let $B$ be a neighborhood of $P$, and compute $H_{*}(X, B) instead), but I suspect that the cut-and-paste stuff will be simpler.