I have the following question:
Suppose a compact $\mathbb{R}$-manifold has finite cell decomposition with only even dimensional cells. Then $M$ is orientable.
It's a theorem that a closed connected $n$ manifold is orientable iff its top homology group $H_n(M;\mathbb{Z})=\mathbb{Z}$.
The cell complex looks something like $$ 0\to\mathbb{Z}^{k_n}\to 0\to\cdots $$ where $\mathbb{Z}^{k_n}$ is the degree $n$-chain group, if there are $k_n$ $n$-cells in $M$. Clearly $H_n(M)=\mathbb{Z}^{k_n}$.
I think the question is ambiguous, and there is the implicit assumption that there is just one cell in each even dimension? Moreover, it seems sufficient that there be a unique cell of highest degree, and no cells of degree one lesser for a manifold to be orientable.
Is this question poorly written, or am I interpreting it wrong?, because my interpretation makes it seem trivial.