Given two compact orientable connected manifolds $M,N$ of dimension $n$ without boundary, and a continuous (or smooth) map $f:M\to N$, the degree of $f$ is defined as the integer
$$
f_*([M])=(\deg f)[N]\in H_n(N)\cong\mathbb{Z}
$$
where $[M]$ is the fundamental class of $M$.
Suppose $\deg f\neq0$, is it possible to show that $\beta_M(q)\geq\beta_N(q)$ for any $1<q<n$, where $\beta_M(q)=\mathrm{rank}~H_q(M;\mathbb{Z})$ is the $q$-th betti number.
I'm inspired by this question and this proof actually shows that $\beta_M(1)\geq\beta_N(1)$ (by the proof the degree is a finite number so the map on $H_1$ becomes an isomorphism after $\otimes_\mathbb{Z}\mathbb{Q}$, and therefore it is a rational surjection). Therefore for surfaces the betti number inequality holds. However, I cannot prove or disprove it.
Best Answer
Yes, this is true.
To see this, notice first that $\beta_m(q) = \dim_\mathbb{Q} H^q(M;\mathbb{Q})$. This follows from the fact that the cohomology groups of $M$ are finitely generated abelian groups.
So, we may as well work with rational cohomology.
Now, given $f:M\rightarrow N$ of non-zero degree, $f$ induces a $(\mathbb{Q}$-linear) map $f^\ast:H^q(N;\mathbb{Q})\rightarrow H^q(M;\mathbb{Q})$. Our goal is to show that this map is injective.
So, select an element $x\in\ker f^\ast$ and assume for a contradiction that $x\neq 0$. Then there must be an element $y\in H^{\dim N - q}(N;\mathbb{Q})$ with $xy\neq 0$; this is a consequence of Poincaré duality (see Hatcher's Algebraic Topology book, Corollary 3.39)
Since $xy\neq 0$ and $f$ has non-zero degree, $f^\ast(xy)\neq 0$. But since $f^\ast(xy) = f^\ast(x)f^\ast(y) = 0 f^\ast(y) = 0$, we have reached a contradiction.