$Ext_{\mathbb{P}^1}^1(F, \mathcal{O}_{\mathbb{P}^1})=H^1(\mathbb{P}^1, F^* \otimes \mathcal{O}_{\mathbb{P}^1})$

My question refers to a step in the proof of THM 2.1.1 (Grotherdieck) from excerpt from "Vector Bundles on Complex Projective Spaces" by Christian Okonek, Michael Schneider, Heinz Spindler (page 12):

Why does hold

$$Ext_{\mathbb{P}^1}^1(F, \mathcal{O}_{\mathbb{P}^1})=H^1(\mathbb{P}^1, F^* \otimes \mathcal{O}_{\mathbb{P}^1})$$

I tried to proof a statement that had been instantly provided wished equality, but the statement was wrong, so I wasn't able to apply it.

How to show the statement above otherwise?

For any locally free sheaf $E$ and any quasi-coherent sheaf $F$ over some scheme $X$, we have $Hom_X(E,F)\cong Hom_X(\mathcal{O}_X,E^\wedge\otimes F)$ where $E^\wedge$ denotes the dual sheaf of $E$. Taking derived functors of both sides and noting that $Hom(\mathcal{O}_X,-)=\Gamma(X,-)$, we obtain the result.

• The point being that an injective resolution $I^\cdot$ of $F$ gives an injective resolution $E^\wedge\otimes I^\cdot$ of $E^\wedge\otimes F$. Correct? – peter a g Apr 20 '18 at 1:45
• Also, we don't need $F$ to be locally free... – peter a g Apr 20 '18 at 2:33
• Yes, your comments are correct. – KReiser Apr 20 '18 at 3:10
• Under which conditions does this statement hold: $I$ injective resolution of $F$ then $E^\wedge\otimes I^\cdot$ is an inj resolution of $E^\wedge\otimes F$? Should $E^\wedge$ have special properties? Locally freeness? – KarlPeter Apr 20 '18 at 11:13
• @KReiser - we don't even need $F$ to be q-c... – peter a g Apr 20 '18 at 12:29