Closed orientable $n$-manifold $X$, there's a map $f: S^n \to X$ of nonzero degree, $n > 1$, is $\pi_1(X)$ finite?

Question

A closed orientable $n$-manifold $X$ satisfies $(*)$ if there is some map $f: S^n \to X$ of nonzero degree (i.e. for which the image of the generator of $H_n(S^n)$ is equal to a nonzero multiple of the generator of $H_n(X)$). If $X$ satisfies $(*)$ and $n > 1$, does it follow that $\pi_1(X)$ is finite?

Answer 1

Let $Y$ be the universal cover of $X$. Then since $n>1$ the map $f$ has a lift $\bar{f}$ to the universal cover $Y$. If $\pi_1(X)$ is infinite, then $Y$ would be non-compact. And $f=p\circ \bar{f}$ where $p$ be the universal covering map. And since the map $f$ is factored through $Y$, so $f_*:H_n(S^n)\to H_n(X)$ will be factored through $H_n(Y)$ which is $0$ (since $Y$ is non-compact, the $n-th$ homology of $Y$ is zero by the version of Poincare Duality of non-compact space). This implies $deg(f)=0$.

Thus if $deg(f)\neq 0$, then $\pi_1(X)$ has to be finite.