Prove or disprove: for every nonzero $\phi \in H^k(M),$ there is $\psi \in H^{n-k}(M)$ with $\phi \cup \psi \neq 0$ ...for $M$ a compact, connected, orientable manifold without boundary.
This is part of a question in a past paper I'm working on: the full thing is at https://www1.maths.ox.ac.uk/system/files/attachments/C3_1_2015.pdf, in question 2. I've tried finding a low-dimensional counterexample: from what I remember, the statement holds for tori of any genus (ruling out a 2-dimensional example). I then tried to find a 3-fold with perhaps a torsion $H_1$, so that $H^2$, isomorphic to $ H_1^{free} \oplus H_0^{tor}$ by the universal coefficient theorem and Poincaré duality, would be zero; and then $H^1 \cup H^2$ would definitely be zero. But I can't find an example, so I'm starting to think it's true...?
This is all $\mathbb{Z}$ cohomology, by the way! Huge thanks in advance to anyone that can find a simple counterexample or a proof at the level of a first course in algebraic topology.
 A: Consider $\mathbb{RP}^3$, the real projective space. It is a compact, connected, orientable manifold of dimension $3$. Its cohomology is given by:
$$H^p(\mathbb{RP}^3;\mathbb{Z}) = \begin{cases}
\mathbb{Z} & p = 0 \\
0 & p = 1 \\
\mathbb{Z}/2\mathbb{Z} & p = 2 \\
\mathbb{Z} & p = 3 \\
0 & p > 3
\end{cases}$$
Then for the nonzero element $\alpha \in H^2(\mathbb{RP}^3;\mathbb{Z})$, there is no element in $H^1$ which multiplies with $\alpha$ to a nonzero element – simply because there's nothing in $H^1$.
As you guessed the problem is indeed torsion: if we consider (co)homology with coefficients in a field $\Bbbk$ instead, let $0 \neq \alpha \in H^k(M; \Bbbk)$. By Poincaré duality, $H^k(M;\Bbbk) \cong H_{n-k}(M;\Bbbk)$, and by the UCT $H_{n-k}(M;\mathbb{Q}) \cong H^{n-k}(M;\Bbbk)$ ($M$ is compact thus $H_{n-k}(M;\Bbbk)$ is a finite-dimensional vector space). Let $\beta \in H^{n-k}(M;\Bbbk)$ correspond to the Poincaré dual of $\alpha$, then $\alpha \cup \beta$ is the fundamental class of $M$ (and hence nonzero), pretty much by definition of Poincaré duality and the UCT isomorphism.
