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It follows from a theorem of Schoen and Yau and the geometrization theorem that $Y^3\times S^1$ admits a metric of positive scalar curvature iff $Y$ does.

On p. 9 of the paper, they define $C_3'$ to be the class of closed 3-manifolds three dimensional manifolds which do not admit any non-zero degree map to a compact three dimensional manifold $M$ such that $\pi_2(M) = 0$ and $M$ contains a two-sided incompressible surface with genus $\geq 1$ (i.e. a Haken 3-manifold).

Let $C_4'$ be the class of closed 4-manifolds $M$ such that every codimension one homology class of $M$ can be represented, up to some non-zero integer, by a map from a manifold of class $C'_3$.

Then Theorem 2 implies that if $M^4$ admits a metric of positive scalar curvature, then $M$ is of class $C'_4$.

Now, consider the 4-manifold $Y^3\times S^1$ and the homology class $[Y^3\times \ast] \in H_3(Y\times S^1)$. It follows from the geometrization theorem and virtual Haken theorem that $Y$ admits a metric of positive scalar curvature iff every finite-sheeted cover $Y'\to Y$ has $Y'\in C_3'$.

Any $M\subset Y'\times S^1$ such that $[M]=[Y'\times \ast]\in H_3(Y'\times S^1)$ is of class $C_3'$. However, there is a non-zero degree map $M\to Y$, hence $Y'\in C_3'$. So we conclude that $Y\times S^1$ admits a psc metric iff $Y$ does.

It follows from Theorem 2 of the following paper and the geometrization theorem that $Y^3\times S^1$ admits a metric of positive scalar curvature iff $Y$ does.

Schoen, Richard; Yau, Shing-Tung, On the structure of manifolds with positive scalar curvature, Manuscr. Math. 28, 159-183 (1979). ZBL0423.53032.

On p. 9 of the paper, they define $C_3'$ to be the class of closed three dimensional manifolds which do not admit any non-zero degree map to a compact three dimensional manifold $M$ such that $\pi_2(M) = 0$ and $M$ contains a two-sided incompressible surface with genus $\geq 1$ (i.e. a Haken 3-manifold). It follows from the geometrization theorem and virtual Haken theorem that $Y$ admits a metric of positive scalar curvature iff every finite-sheeted cover $Y'\to Y$ has $Y'\in C_3'$.

Let $C_4'$ be the class of closed 4-manifolds $M$ such that every codimension one homology class of $M$ can be represented, up to some non-zero integer, by a map from a manifold of class $C'_3$ (a homology class $\alpha\in H_k(M)$ is represented by an oriented $k$-manifold $N$ if there is a map $f:N\to M$ such that $f_\ast([N])=\alpha$, where $[N]\in H_k(N)$ is the fundamental class determined by the orientation of $N$).

Then Theorem 2 implies that if $M^4$ admits a metric of positive scalar curvature, then $M$ is of class $C'_4$.

Now, consider the 4-manifold $Y^3\times S^1$ and let $Y'\to Y$ be a finite-sheeted cover. By Theorem 2, there exists an $N\subset Y'\times S^1$ such that $[N]=[Y'\times \ast]\in H_3(Y'\times S^1)$ and $N$ is of class $C_3'$. Moreover, there is a non-zero degree map $N\to Y'$ (by composing the embedding with the projection $Y'\times S^1\to Y'$), hence $Y'\in C_3'$ (the class $C_3'$ is clearly closed under taking non-zero degree maps). Since every finite-sheeted cover of $Y$ is in $C_3'$, we conclude that $Y$ admits a metric of positive scalar curvature.

It follows from a theorem of Schoen and Yau and the geometrization theorem that $Y^3\times S^1$ admits a metric of positive scalar curvature iff $Y$ does.

On p. 9 of the paper, they define $C_3'$ to be the class of closed 3-manifolds three dimensional manifolds which do not admit any non-zero degree map to a compact three dimensional manifold $M$ such that $\pi_2(M) = 0$ and $M$ contains a two-sided incompressible surface with genus $> 1$ (i.e. a Haken 3-manifold).

Let $C_4'$ be the class of closed 4-manifolds $M$ such that every codimension one homology class of $M$ can be represented, up to some non-zero integer, by a map from a manifold of class $C'_3$.

Then Theorem 2 implies that if $M^4$ admits a metric of positive scalar curvature, then $M$ is of class $C'_4$.

Now, consider the 4-manifold $Y^3\times S^1$ and the homology class $[Y^3\times \ast] \in H_3(Y\times S^1)$. It follows from the geometrization theorem and virtual Haken theorem that $Y$ admits a metric of positive scalar curvature iff every finite-sheeted cover $Y'\to Y$ has $Y'\in C_3'$.

Any $M\subset Y'\times S^1$ such that $[M]=[Y'\times \ast]\in H_3(Y'\times S^1)$ is of class $C_3'$. However, there is a non-zero degree map $M\to Y$, hence $Y'\in C_3'$. So we conclude that $Y\times S^1$ admits a psc metric iff $Y$ does.

It follows from a theorem of Schoen and Yau and the geometrization theorem that $Y^3\times S^1$ admits a metric of positive scalar curvature iff $Y$ does.

On p. 9 of the paper, they define $C_3'$ to be the class of closed 3-manifolds three dimensional manifolds which do not admit any non-zero degree map to a compact three dimensional manifold $M$ such that $\pi_2(M) = 0$ and $M$ contains a two-sided incompressible surface with genus $\geq 1$ (i.e. a Haken 3-manifold).

Let $C_4'$ be the class of closed 4-manifolds $M$ such that every codimension one homology class of $M$ can be represented, up to some non-zero integer, by a map from a manifold of class $C'_3$.

Then Theorem 2 implies that if $M^4$ admits a metric of positive scalar curvature, then $M$ is of class $C'_4$.

Now, consider the 4-manifold $Y^3\times S^1$ and the homology class $[Y^3\times \ast] \in H_3(Y\times S^1)$. It follows from the geometrization theorem and virtual Haken theorem that $Y$ admits a metric of positive scalar curvature iff every finite-sheeted cover $Y'\to Y$ has $Y'\in C_3'$.

Any $M\subset Y'\times S^1$ such that $[M]=[Y'\times \ast]\in H_3(Y'\times S^1)$ is of class $C_3'$. However, there is a non-zero degree map $M\to Y$, hence $Y'\in C_3'$. So we conclude that $Y\times S^1$ admits a psc metric iff $Y$ does.

Schoen and Yau define a class $\mathcal{C}_n$ of closed smooth $n$-manifolds inductively. $\mathcal{C}_3$ consists of 3-manifolds not containing a surface subgroup. By the geometrization theorem and surface subgroup theorem, this is equivalent to the class of 3-manifolds of positive scalar curvature (equivalently, those which have virtually free fundamental group). Then $\mathcal{C}_4$ is the class of manifolds such that for every covering space, any codimension-1 homology class is realized by a manifold in $\mathcal{C}_3$ or the $\hat{A}$-genus vanishes (which is implied by positive scalar curvature). They prove that a 4-manifold with positive scalar curvature is of class $\mathcal{C}_4$.

In the case at hand of $Y\times S^1$, $[Y\times \ast]\in H_3(Y\times S^1)$ is a codimension-one homology class, hence must be represented by a 3-manifold in $\mathcal{C}_3$ if $Y\times S^1$ has positive scalar curvature. But if $M\subset Y\times S^1, [M]=[Y]\in H_3(Y\times S^1)$, then there is a non-zero degree map $M\to Y$, and hence if $M\in \mathcal{C}_3$, then $Y\in \mathcal{C}_3$. Hence $Y\times S^1$ admits a metric of positive scalar curvature iff $Y$ does.

It follows from a theorem of Schoen and Yau and the geometrization theorem that $Y^3\times S^1$ admits a metric of positive scalar curvature iff $Y$ does.

On p. 9 of the paper, they define $C_3'$ to be the class of closed 3-manifolds three dimensional manifolds which do not admit any non-zero degree map to a compact three dimensional manifold $M$ such that $\pi_2(M) = 0$ and $M$ contains a two-sided incompressible surface with genus $> 1$ (i.e. a Haken 3-manifold).

Let $C_4'$ be the class of closed 4-manifolds $M$ such that every codimension one homology class of $M$ can be represented, up to some non-zero integer, by a map from a manifold of class $C'_3$.

Then Theorem 2 implies that if $M^4$ admits a metric of positive scalar curvature, then $M$ is of class $C'_4$.

Now, consider the 4-manifold $Y^3\times S^1$ and the homology class $[Y^3\times \ast] \in H_3(Y\times S^1)$. It follows from the geometrization theorem and virtual Haken theorem that $Y$ admits a metric of positive scalar curvature iff every finite-sheeted cover $Y'\to Y$ has $Y'\in C_3'$.

Any $M\subset Y'\times S^1$ such that $[M]=[Y'\times \ast]\in H_3(Y'\times S^1)$ is of class $C_3'$. However, there is a non-zero degree map $M\to Y$, hence $Y'\in C_3'$. So we conclude that $Y\times S^1$ admits a psc metric iff $Y$ does.