Let $h$ be the moduli functor $(\text{Schemes}_{\mathbb{C}})^{op}\to \text{Sets}$ described as follows.
On objects $h(X)$ gives us the following: a line bundle $L\to X$ with an injective map of vector bundles $i:L \to X\times \mathbb{C}^{n+1}$ with a quotient that is a vector bundle (i.e. a constant rank vector bundle map).
On morphisms, we apply the pullback the line bundle and its vector bundle map.
Now, we want to show that this moduli functor $h$ is representable and is naturally isomorphic to $h^{\mathbb{P}^n}$ .
$h(X)$ gives us a short exact sequence of locally free sheaves, $0\to \mathcal{L}\to \mathcal{O}_X^{n+1}\to \mathcal{Q}\to 0$ where $\mathcal{L}$ is the sheaf associated to the line bundle $L$ and $Q$ is the quotient bundle with a locally free sheaf $\mathcal{Q}$ associated to it.
This is where I start to lose track of what's going on.
"Taking the dual, this is equivalent to an exact sequence,
$0\to \mathcal{Q}^{\vee}\to \mathcal{O}_X^{n+1}\to \mathcal{L}^{\vee}\to 0$
of locally free sheaves. Since the kernel of a map of locally free sheaves is automatically locally free, this is equivalent to just specifying the surjection $\iota^{\vee}: \mathcal{O}_X^{n+1}\to \mathcal{L}^{\vee}$. Such a surjection is determined by specifying sections $s_0,...,s_n\in H^0(X,\mathcal{L}^{\vee}) $ without a common zero...."
Question 1: Why is taking duals exact? In particular, why is it right exact?
Question 2: The idea is that we want to find $n+1$ linearly independent basepoint-free sections of $\mathcal{L^{\vee}}$ (or perhaps it should be $\mathcal{L}$?) to construct a map $X\to \mathbb{P}^n$ (that's what $h^{\mathbb{P}^n}(X)$ is). But taking global sections is only left-exact so $H^0(X,\mathcal{O}_X^n)\to H^0(X,\mathcal{L}^{\vee})$ need not be a surjection of vector spaces (and even if it were we'd still need that $H^0(X,\mathcal{Q}^{\vee})$ to be exactly $0$). How are we concluding that $H^0(X,\mathcal{L}^{\vee})$ has dimension $\geq n+1$?