The matrix in part (d) has trace $4$ and determinant $1$. Since $|\operatorname{tr}\phi_*|>2$, the once-punctured-torus case of the <Nielsen–Thurston classification theorem> says that $\phi$ is pseudo-Anosov. The <Hyperbolization of a pseudo-Anosov mapping torus> therefore gives
$$
\boxed{\operatorname{int}M_{\mathcal H}\text{ is a complete finite-volume hyperbolic three-manifold}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 141 / 2 / e / Solution

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{wiki=Nielsen–Thurston_classification}

The Nielsen–Thurston classification says that an orientation-preserving surface homeomorphism is periodic, reducible, or pseudo-Anosov. For a once-punctured torus, the three cases correspond to $|\operatorname{tr}M|<2$, $|\operatorname{tr}M|=2$, and $|\operatorname{tr}M|>2$ for its action $M\in SL_2(\mathbb Z)$ on first homology.

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 Incoming links: Nielsen–Thurston classification theorem