LS category of 4-manifolds with free fundamental group

Question

In this math overflow question asked almost 5 years ago about the (normalized) Lusternik-Schnirelmann category of $4$-manifolds, the second-last comment (by Jeff Strom) says the following:

A $4$-dimensional compact manifold with free $\pi_1$ has L-S category $2$.

Specifically, if $\pi_1(M) = 0$ and $M \ne S^4$ is a $4$-manifold, then the above is true (as remarked by Mark Grant in the above question post). But I am not sure why the above statement is true for any $4$-manifold $M$ with $\pi_1(M)$ free and non-trivial. As observed by Oprea–Strom, it can be deduced from the work of Matumoto–Katanaga (and Hellen Colman) that the LS category of such manifolds is at most $2$. I don't understand why their LS category should be exactly $2$ (in particular, why can't it be $1$). Can someone please provide a reason/reference?

Answer 1

Since we have settled on an argument in the comments, let me post it as an answer.

We have to show that a closed $4$-manifold with nontrivial free $\pi_1(M)$ does not have $\mathrm{cat}(M)=1$. Indeed, if $\pi_1(M)$ is free, $H_1(M)$ is free abelian, and so in particular $H^1(M;R)$ with any coefficients is nontrival by the universal coefficient theorem. One version of Poincaré duality [Hatcher, Algebraic Topology, Proposition 3.38] says that for closed $M$, the pairing $$ H^i(M;\mathbb{F}_2) \otimes H^{n-i}(M;\mathbb{F}_2) \to H^n(M;\mathbb{F}_2)\cong \mathbb{F}_2 $$ is nondegenerate. In the case at hand, this means that for any nontrivial element of $H^1(M;\mathbb{F}_2)$, we find one in $H^3(M;\mathbb{F}_2)$ such that their cup product is nonzero. So $M$ has cup length at least $2$, and thus $\mathrm{cat}(M)\geq 2$.

In the comments, Tyrone mentioned a stronger result along those lines, namely that a closed connected manifold (or more generally Poincaré complex) with $\mathrm{cat}(M)=1$ must be a homology sphere. Indeed, in dimension $n=1$ there is nothing to show, and otherwise the above argument shows that $H^1(M;\mathbb{F}_2)=0$. But then $M$ is orientable, and so we have Poincaré duality with arbitrary coefficients and conclude that $H^i(M;K)=0$ for any $0<i<n$ and any field $K$. By a standard argument (using $K=\mathbb{F}_p$ and $\mathbb{Q}$) this shows that also the integral cohomology vanishes in those degrees, and thus that $M$ is a homology sphere.

(In the comments it was claimed that it suffices for $M$ to be a complex with Poincaré duality only with $\mathbb{F}_2$ coefficients, but I don't see how to argue then, and suspect something like the suspension of a Moore space with homology $\mathbb{Q}_2/\mathbb{Z}_2$ is a counterexample. This "looks like" a homology sphere with $\mathbb{F}_2$-coefficients, but not away from $2$.)