 Solution

For the triangle quotient, the universal coefficient theorem for homology gives $H_j(X;{\mathbb{F}}_2)=0$ for every $j>0$, because its only positive integral homology is $\mathbb{Z}/3$. A connected positive-dimensional closed manifold has a nonzero top mod-two fundamental class, independently of orientability. Thus this quotient cannot be homotopy equivalent to such a manifold. A connected zero-dimensional closed manifold is a point, which its nonzero $H_1(X;\mathbb{Z})$ also excludes.

For the attached-band space, the same universal coefficient theorem for homology gives nonzero $H_2(X;{\mathbb{F}}_2)$ and zero homology above degree two. A connected closed manifold homotopy equivalent to it would consequently have dimension two. Its Euler characteristic would be one. The classification theorem for surfaces says that a closed connected surface of Euler characteristic one is the real projective plane. But its integral $H_1$ is $\mathbb{Z}/2$, differing from both possibilities computed in part 2. Thus this space is never homotopy equivalent to a closed manifold.

The antipodal mapping torus is already a closed manifold: gluing the two boundary components of $S^n\times [0,1]$ by a diffeomorphism gives a smooth $(n+1)$-manifold without boundary. For $n\ge 1$, it is orientable exactly when $(-1{)}^{n+1}=1$, namely when $n$ is odd. In that case $-I$ on ${\mathbb{R}}^{n+1}$ is isotopic to the identity through rotations, so the mapping torus is diffeomorphic to $S^n\times S^1$. The $n=0$ case is a circle as above. Only the third family can be homotopy equivalent to a closed manifold.