If $M_g$ denotes the closed orientable surface of genus $g$, show that degree $1$ maps $M_g\to M_h$ exist iff $g\geq h$.
Construction of degree $1$ map for $g\geq h$ is easy. I want to prove the converse using cup product. From now on, the basic idea is given by @TedShifrin. First I consider $(g,h) = (0,1)$ and $(1,2)$ cases for some observation.
Before the argument, by the naturality of UCT, we have a commutative diagram
$$\require{AMScd} \begin{CD} H^2(M_h) @>\sim>> \operatorname{Hom}(H_2(M_h),\Bbb Z)\\ @Vf^* VV @V(f_*)^*VV\\ H^2(M_g) @>\sim>> \operatorname{Hom}(H_2(M_g),\Bbb Z) \end{CD}$$
So, the degree of $f_*$ defined in terms of the fundamental class is same for $f^*$ i.e., if $M$ is a dual of fundamental class of $H_2(M_h)$, and $N$ for $H_2(M_g)$, $f^*(N) = d\cdot M$ where $d$ is a degree of $f$.
Also, I use the notation $\alpha_i,\beta_i$ for the generators of $H^1(M_h)$ and $\alpha_i',\beta_i'$ for $H^1(M_g)$ (See Example 3.7 in Hatcher for detail). $\gamma$ is a generator of $H^2(M_h)$ and $\gamma'$ for $H^2(M_g)$.
Actually for $(g,h) = (0,1)$ case, the argument is easy. Suppose $f$ is a degree $1$ map. Then $f^*(\gamma) = f^*(\alpha_1\smile\beta_1) = f^*(\alpha_1)\smile f^*(\beta_1)$. By assumption, $f^*(\gamma) =\gamma'$ but as $M_0 = S^2$ has vanishing homology at degree $1$, $f^*(\alpha_1) = f^*(\beta_1) = 0$. Hence, $f^*(\alpha_1)\smile f^*(\beta_1) =0$ which is a contradiction.
Now consider the case $(g,h) = (1,2)$. We have $f^*(\alpha_1) = a_{11}\alpha'_1+b_{11}\beta_1'$, $f^*(\alpha_2) = a_{21}\alpha_1'+b_{21}\beta'_1$, $f^*(\beta_1)= a'_{11}\alpha'_1+b'_{11}\beta_1'$ and $f^*(\beta_2) = a'_{21}\alpha'_1+b'_{21}\beta'_1$. Computation shows that
\begin{align*}
& f^*(\alpha_1)\smile f^*(\beta_1) = (a_{11}b'_{11}-a'_{11}b_{11})\gamma' = \gamma'\\
& f^*(\alpha_2)\smile f^*(\beta_2) = (a_{21}b'_{21}-a'_{21}b_{21})\gamma' = \gamma'\\
& f^*(\alpha_1)\smile f^*(\alpha_2) = (a_{11}b_{21}-b_{11}a_{21})\gamma' = 0\\
& f^*(\beta_1)\smile f^*(\beta_2) = (a'_{11}b'_{21}-a'_{21}b'_{11})\gamma' = 0\\
& f^*(\alpha_1)\smile f^*(\beta_2) = (a_{11}b'_{21}-a'_{21}b_{11})\gamma' = 0\\
& f^*(\alpha_2)\smile f^*(\beta_1) = (a_{21}b'_{11}-a'_{11}b_{21})\gamma' = 0\\
\end{align*}
Well, the above computation shows that $f^*(\alpha_1)$ and $f^*(\beta_1)$ are $\Bbb Z$-linearly independent (determinant). But $f^*(\alpha_2)$ and $f^*(\alpha_1)$ are $\Bbb Z$-linearly dependent and $f^*(\alpha_2)$ and $f^*(\beta_1)$ are $\Bbb Z$-linearly dependent which is a contradiction.
I wonder this method (especially the latter one) can be generalized for general $g<h$. Or is there other approach to this?
Best Answer
One can use Poincaré duality to show much more.
Lemma: 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 non-zero degree, and $x \in H^k(M_2; \mathbb{Z})$ is non-torsion, then $f^*x \neq 0$.
Proof: Let $\alpha_i$ denote the oriented generator of $H^n(M_i; \mathbb{Z})$; note that $f^*\alpha_2 = \deg(f)\alpha_1$. As $x \in H^k(M_2; \mathbb{Z})$ is non-torsion, by Poincaré duality there is $y \in H^{n-k}(M_2; \mathbb{Z})$ with $x\cup y\neq 0$, so $x\cup y = r\alpha_2$ for some $r\neq 0$. Now
$$f^*x\cup f^*y = f^*(x\cup y) = f^*(r\alpha_2) = rf^*\alpha_2 = r\deg(f)\alpha_1 \neq 0,$$ so $f^*x \neq 0$. $\qquad\square$
That is, the kernel of the map $f^* : H^k(M_2; \mathbb{Z}) \to H^k(M_1; \mathbb{Z})$ is contained in the torsion subgroup of $H^k(M_2; \mathbb{Z})$, so $f^* : H^k(M_2; \mathbb{Z})/\text{torsion} \to H^k(M_1; \mathbb{Z})$ is injective.
Corollary: If there exists a map $M_1 \to M_2$ of non-zero degree, the Betti numbers satisfy $b_k(M_1) \geq b_k(M_2)$ for all $k$.
In particular, for $g < h$, there is no map $\Sigma_g \to \Sigma_h$ of non-zero degree.