We work over an algebraically closed field. Suppose $X\subset \mathbf{P}^n$ is an integral projective curve and $\pi:X\to Y$ is a linear projection that identifies two distinct points $p,q\in X$ to a point $y\in Y$ and is an isomorphism elsewhere.
I want to show that $h^0(Y,\mathscr{O}_Y(n)) = h^0(X,\mathscr{O}_X(n))-1$ for all $n\ge 1$.
Here is my proof: The natural map $H^0(Y,\mathscr{O}_Y(n)) \xrightarrow{f} H^0(X,\pi^*\mathscr{O}_Y(n))$ is injective and $\pi^*\mathscr{O}_Y(n) = \mathscr{O}_X(n)$. Consider $H^0(X,\mathscr{O}_X(n)) \xrightarrow{g} k$ that maps $\sigma$ to $\sigma(p)-\sigma(q)$, where $\sigma(p)$ is the image of $\sigma_p$ in $k$ under a choice of isomorphism $(\pi^*\mathscr{L})_p/m_p(\pi^*\mathscr{L})_p \cong k$, and the same for $q$. All such choices change $\sigma(p)$ (and $\sigma(q)$) by a nonzero scalar. Since $\mathscr{O}_X(n)$ is very ample for $n\ge 1$, we may find a section that vanishes on $p$ and not on $q$. This means that $g$ is surjective for $n\ge 1$, and clearly the image of $f$ is in the kernel of $g$. Thus we see that $h^0(Y,\mathscr{O}_Y(n))\le h^0(\mathscr{O}_X(n))-1$. To prove equality, consider the sequence $$0\to \mathscr{O}_Y \to \pi_* \mathscr{O}_X \to k(y) \to 0.$$ The sequence is exact on the left because $\pi$ is surjective, and the cokernel is supported on $y$, and it has length $1$, thus is $k(y)$. Twisting by $\mathscr{O}_Y(n)$, and taking cohomology, we see that $$h^0(X,(\pi_*\mathscr{O}_X)\otimes \mathscr{O}_Y(n))-h^0(Y,\mathscr{O}_Y(n))\le 1.$$ However, by projection formula $$\pi_*(\mathscr{O}_X(n)) \cong \pi_*(\mathscr{O}_X\otimes \pi^*\mathscr{O}_Y(n)) \cong (\pi_*\mathscr{O}_X)\otimes \mathscr{O}_Y(n)$$ and since $\pi$ is finite we have $R^i\pi_* = 0$ for $i>0$ and
$$H^0(X,\mathscr{O}_X(n))\cong H^0(Y,\pi_*\mathscr{O}_X(n))\cong H^0(Y,(\pi_*\mathscr{O}_X)\otimes \mathscr{O}_Y(n)).$$ Therefore we have equality.
I have two questions: (1) Is the proof correct? I feel a little strange about it because I seem to draw weird consequences from this. (2) Is there a direct way to show that any section of $H^0(X,\mathscr{O}_X(n))$ that agrees on $p$ and $q$ must come from a section in $H^0(Y,\mathscr{O}_Y(n))$? What's the best way to analyse $H^0(Y,\mathscr{F})\to H^0(X,\pi^*\mathscr{F})$ in this situation?
