Does there exist $f:M \rightarrow S^n$ s.t. $f_{*}([M])=[S^n]$?

Question

Let $M$ be an orientable n-dimensional manifold, $[M]\in H_{n}(M) \cong Z$ be the fundamental class of $M$, and $[S^n]\in H_{n}(S^n)$ be the fundamental class of $S^n$.
I want to know how to prove there exists a mapping $f:M \rightarrow S^n$ such that $f_{*}([M])=[S^n]$.

Answer 1

Let $U \subseteq M$ be an open subset of $M$ homeomorphic to $\mathbb{R}^n$. Let $V \subseteq U$ be the homeomorphic image of the unit disk (so that the closure of $V$ is inside $U$). Define a continuous $f: M \to S^n$ by mapping $M-V$ to a fixed point $p \in S^n$, and mapping $\bar{V}$ to the sphere via any homeomorphism $\bar{V}/\partial V \to S^n$ which maps the boundary of $V$ to $p$.

To prove that the pushforward of a fundamental class is actually the fundamental class, observe that $f: (M, \ast) \to (S^n, p)$ (for $\ast$ any point outside $V$) factors through $(M, M-V)$. Since $M-V$ is an $n$-manifold with boundary, $\tilde{H}_n(M-V) = 0$, so $\tilde{H}_n(M) \to H_n(M, M-V)$ is injective. Since the map from $\tilde{H}_n(M) \to H_n(M,M-q)$ factors through this map for any $q \in V$, and the latter is an isomorphism, we see that the image of the fundamental class generates a direct summand of $\tilde H_n(M, M-V)$. By excision, $H_n(M, M-V) \cong H_n(U, U-V)$, and the map $(U, U-V) \to (S^n, p)$ is an isomorphism on homology. Since the isomorphism by excision is induced by the inclusion $(U, U-V) \to (M, M-V)$, the map $f$ induces on $(M, M-V) \to (S^n, p)$ is an isomorphism on homology. So the image of the fundamental class of $[M]$ under $M \to (M, M-V)$ is the generator of $\tilde{H}_n(M, M-V)$, and hence $f_\ast[M]$ is the fundamental class of the sphere.