Showing that $S^2 \vee S^4$ is not homotopy equivalent to any closed 4-manifolds. I am working on a problem that asks to show that $S^2 \vee S^4$ is not homotopy equivalent to any closed $4$-manifold. My only understanding of manifolds is from Hatcher's chapter on Cohomology (chapter 3) so all manifolds in question are general topological manifolds. 
From what I was told, I should consider working with $F_2$ coefficients and using poincare duality. However, I do not see what to do.
 A: The key point is that Poincare duality is given by capping against the fundamental class. 
You can write down the cohomology ring of S2∨S4 (wedge sum of spaces), where it should be clear that all products are zero, and conclude that when you cap the fundamental class against a generator for the cohomology in degree 2, you get zero, which is not what would happen under Poincare duality...
Assume for contradiction that Poincare duality holds for $S^2 \vee S^4$. In particular, we let $\phi$ generate the second cohomology, $[M]$ be the fundamental class (generator of top homology), and let $\sigma$ be a generator of second homology. 
Assuming it has Poincare duality implies that $\phi \cap [M] = \sigma$ (we're using $F_2$ coefficients).
However, because the general formula $(\alpha \cap \phi, \psi) = (\alpha, \phi \cup \psi)$ holds ( $(,)$ denotes the pairing betwee homology and cohomology... see the Hatcher document below), we have $1 = (\sigma, \phi) = ([M] \cap \phi, \phi) = ([M], \phi \cup \phi) = ([M], 0) = 0$, which is a contradiction.
math.cornell.edu/~hatcher/AT/ATcapprod.pdf gives a helpful discussion
