 Solution

Put $N=S^d/X$ and let $p=q(X)$. The restrictionis a homeomorphism. At every point of $N\setminus \{p\}$, transport the local orientation of $S^d$ through this homeomorphism. The localization of the global class $q_{\ast }[S^d]$ is therefore a generator at one, and hence every, point of that connected open set. The localizations of a global homology class form a section of the orientation local system; because the manifold $N$ is connected, this generator extends across $p$. Thus $q_{\ast }[S^d]$ is a fundamental class for an orientation of $N$.

Choose $[N]=q_{\ast }[S^d]$. By the degree of a map between oriented manifolds,so $\boxed{\mathrm{deg}q=1}$.