Let $\Gamma\subset\mathbb{P}^2$ be a singular plane curve and let us consider $f:C\to\Gamma$ the normalization map. Given $L\in\mathrm{Pic}(\Gamma)$, is it always true that $h^0(\Gamma,L)=h^0(C, f^\*L)$? If not, can you exhibit a counterexample?



Pull-back of line bundles under the normalization map.
Alternately, for any such $\Gamma$ and $C$, we always have a short exact sequence $$0 \to O\_{\Gamma} \to f\_\* O\_C \to F \to 0,$$
where $F$ is a finite length $\mathcal{O}\_{\Gamma}$-module supported where $\Gamma$ is non-smooth. 


Twisting by a very positive (ie, high multiple of an ample) Cartier divisor $L$ and taking cohomology we get
$$
0 \to H^0(\Gamma, O\_{\Gamma}(L)) \to H^0(C, O\_{C}(f^\*L)) \to H^0(\Gamma, F(L)) \to H^1(\Gamma, O\_{\Gamma}(L))$$


Now, $H^0(\Gamma, F(L)) \neq 0$ but $H^1(\Gamma, O\_{\Gamma}(L)) = 0$ by Serre vanishing. And so you have a counter-example.


