I want to apologize first of all if my English is bad, I hope my message will still be understandable.
I’m currently reading Lazarsfeld’s book "Positivity in Algebraic Geometry I" and I’m stuck on one obvious point of Theorem 1.8.3 (p.99). $V$ is a complex vector space of dimension $r+1$, $\mathbb{P} = \mathbb{P}(V^\vee)$, $V_{\mathbb{P}} := V \otimes \mathcal{O}_{\mathbb{P}}$ and $\mathcal{F}$ is a $m$-regular (in the sense of Castelnuovo-Mumford) coherent sheaf on $\mathbb{P}$. I don’t think we need all that for my question, which seems to have a very general answer. In one place of the proof, Lazarsfeld writes :
$$H^0(\mathbb{P},\mathcal{F}(m)) \otimes H^0(\mathbb{P}, \mathcal{O}_{\mathbb{P}}(1)) = H^0(\mathbb{P}, V_{\mathbb{P}} \otimes \mathcal{F}(m))$$
Why do we have this equality (isomorphism)? I have many shortcomings that I fill little by little but I remain stuck on this point. Do you have a lead to help me ? The rest of the proof is clear to me, I’m sorry if this question is obvious.Thank you very much.
In the proof in the mentioned book, we start with $V$ being the $r + 1$-dimensional vector space used to define $\mathbb{P} = \mathbb{P}(V^\vee) \cong \mathbb{P}^r$, and identify it with $H^0(\mathbb{P}, \mathcal{O}(1))$ by starting with a surjection $V \otimes \mathcal{O}(-1) \to \mathcal{O}$. Indeed, twisting by 1 we get a surjection $V \otimes \mathcal{O} \to \mathcal{O}(1)$, which we claim remains a surjection when we take $H^0$.
Since this map is a surjection, it follows that its image $W$ defines a free linear series $|W|$ contained in the complete series $|\mathcal{O}(1)|$. If the dimension of $W$ is less than $r + 1$, then $W$ defines a morphism $\mathbb{P}^r \to \mathbb{P}^\ell$ for $\ell < r$. Hartshorne's exercise II.7.3 tells us this cannot be.
Hence, $V = H^0(V \otimes \mathcal{O}) \to H^0(\mathcal{O}(1))$ is a vector space isomorphism. As such, going forward, we may assume that $V = H^0(\mathcal{O}(1))$ and the map $V \otimes \mathcal{O}(-1) \to \mathcal{O}$ is the multiplication map map $(s|_U, t) \mapsto s|_U \otimes t$. (Note: using tensor-hom adjunction this is identified with the usual evaluation map $(s, \phi) \mapsto \phi(s)$.)
Remark: As far as I understand, the proof just requires the existence of a surjection $V \otimes \mathcal{O}(-1) \to \mathcal{O}$, and the construction goes from there. As such in the proof we may just choose this map to be the twisting of the multiplication map $V \otimes \mathcal{O} \to \mathcal{O}(1)$ sending basis $e_i$ to variables $x_i$ of $\mathbb{P}^r$. What we observed above, which is that we may identify any surjection which this kind of map, is therefore somewhat unnecessary. I only include it since the book in question does not specify what this map is.
Now, continuing on, in the proof we construct the Koszul complex associated to $V \otimes \mathcal{O}(-1) \to \mathcal{O}$ (which is an exact complex of locally free sheaves, so tensoring with $\mathcal{F}(m)$ preserves the exactness). We claim that the image of
$$H^0(V \otimes \mathcal{F}(m)) \to H^0(\mathcal{F}(m + 1))$$ is exactly the image of the multiplication map $H^0(\mathcal{O}(1)) \otimes H^0(\mathcal{F}(m)) \to H^0(\mathcal{F}(m + 1))$. But this now follows from the identifications we've made. Indeed, this map $V \otimes \mathcal{F}(m) \to \mathcal{F}(m + 1)$ is the tensor product of the multiplication map $V \otimes \mathcal{O} \to \mathcal{O}(1)$ with $\operatorname{Id}_{\mathcal{F}(m)}$, which is exactly the multiplication map $V \otimes \mathcal{F}(m) \to \mathcal{F}(m + 1)$.