= 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 $n\geq1$, 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. \b[Only the third family can be homotopy equivalent to a closed manifold.]
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