I've been reading a book of tessellations and three-manifolds and it's been said that "the fiber of ST(X) is the 1-sphere and thus ST(X) is a closed 3-manifold", where ST(X) is the spherical tangent bundle of a compact, connected 2-manifold X, that is, the subbundle of TX consisting of vectors of norm 1. Honestly, I don't understand why the assertion between quotations marks is true (why it is a 3-manifold). Could anyone help me?
Many thanks.
Fiber bundles $F \to E \to X$ are locally trivializable, meaning there is an open neighborhood $U$ of any $x\in X$ Such that the inverse image of $U$ under the projection is diffeomorphic to $U \times F$, where $F$ is the fiber. In this case, we may choose $U$ to be diffeomorphic to an open neighborhood of $\mathbb{R}^2$. Since the tangent space $V$ to a point $x \in X$ is isomorphic to $\mathbb{R}^2$, we have $F \cong S^1$. This tells us that $ST(X)$ is a $3$-manifold.
As for why it's closed, since $X$ is compact and $ST(X)\to X$ has compact fibers, $ST(X)$ will be compact as well.