Let $f: C \to D$ a dominant morphism which is not an isomorphism between two irreducible, reduced, projective curves $C,D$ over an alg closed field $k$ (unsure if algebraically closedness is neccessary). Let $L$ be line bundle on $D$, $L^n = L^{\otimes n}$ it's $n$-th power and $K^m:= f^*L^m$ it's pullback. Assume that $K^m$ is very ample.
Question: Why $h^0(C, K^m) > h^0(D, L^m)$ is a proper inequality? That is why the case $h^0(C, K^m) = h^0(D, L^m)$ cannot happen if $k^m$ is very ample?
Recall Eoin's argument here shows that $h^0(C, K^m) \ge h^0(D, L^m)$ since $f$ dominant implies that $O_D \to f_*O_C$ is injective and twisting by $L^m$ leads to injective map $L^m \to f_*O_C \otimes L^m= f_*(K^m)$ (the last one is projection formula). Well, why $h^0(C, K^m) > h^0(D, L^m)$.
my idea: $f$ is finite since between curves and since not isomorphism, we can find an affine open subscheme $U \subset D$ such that the restriction of $f$ to affine open $V:=f^{-1}(U) \subset C$ is not an isomorphism. We obtain a injective ring map $\varphi:R_U \to R_V$. The restriction wasn't an isomorphism thus it's not surjective. Twisting with local sections $L^m(U)$ of line bundle preserves the non surjectivity. Thus $\varphi \otimes id_{L^m(U)}: L^m(U) \to K^m(V)$ is still not surjective and we find a $s_V \in K^m(V)$ which haven't a preimage in $L^m(U)$.
Can we use very ampleness of $K^m$ to extend $s_V$ to a global section $s \in H^0(C,K^m)=K^m(C)$?