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게시글 주소: https://orbi.kr/00040027196
Let $M_1$ and $M_2$ be closed oriented manifolds of dimension $n$. If $f:M_1\to M_2$ is a continuous map of nonzero degree and $x\in H^k(M_2;\Bbb Z)$ is non-torsion, then $f^*(x)\neq 0$\\
Proof. Let $[M_i]$ denote the fundamental class of $H^n(M_i;\Bbb Z)$. Note that $f^*([M_2]) = \deg(f)[M_1]$. As $x\in H^k(M_2;\Bbb Z)$ is non-torsion, by Poincare duality, the cup product pairing is nonsingular. Hence, there is $y\in H^{n-k}(M_2;\Bbb Z)$ with $x\smile y\neq 0$. Hence, $x\smile y = r[M_2]$ for some $r\neq 0$. Now we have
$$ f^*(x)\smile f^*(y) = f^*(x\smile y) = f^*(r[M_2]) = rf^*([M_2]) = r\deg(f)[M_1]\neq 0$$
by assumption so $f^*(x)\neq 0$. $\square$\\
In particular, $f^*:H^k(M_2;\Bbb Z)/\text{torsion}\to H^k(M_1;\Bbb Z)$ is injective.\\
Hence if there is a non-zero degree map $M_1\to M_2$ then its Betti number should satisfy $b_k(M_1)\geq b_k(M_2)$ for all $k$.
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