 Solution

For any finite-dimensional bounded cochain complex $(V^{\ast },d)$ over a field, let $r_q=\mathrm{rank}(d:V^q\to V^{q+1})$, with zero ranks outside its degree range. ThenTaking the alternating sum cancels the two rank sums. Thus taking cohomology preserves the Euler characteristic of a finite graded complex. Apply this to $V^q=H^q(M;{\mathbb{F}}_p)$ and $d=\beta$, using part (b).

A closed three-dimensional manifold has finite-dimensional cohomology, vanishing above degree three. Its Euler characteristic is zero, even when it is not orientable: Poincare duality with ${\mathbb{F}}_2$ coefficients gives $b_q=b_{3-q}$, so the alternating sum vanishes. A finite triangulation, or finite CW complex model, shows that the alternating sum is the same integer with coefficients in any field, since it equals the alternating count of cells. ThereforeThe primeness assumption makes $\mathbb{Z}/p$ a field; no orientation assumption on $M$ is needed.