Let $\mathcal{F} \in Coh(\mathbb{P}^1 _k)$ a coherent sheaf then we have the bilinear morphism, also called the "canonical" pairing: $$Hom(\mathcal{O}_{\mathbb{P}^1}, \mathcal{F}) \times Ext^1(\mathcal{F}, \omega_{\mathbb{P}^1}) \to Ext^1(\mathcal{O}_{\mathbb{P}^1}),\omega_{\mathbb{P}^1})$$
where $Ext^1(\mathcal{F},-)$ (resp. $Ext^1(-,\mathcal{F}))$ is the derived functor of $Hom(\mathcal{F},-)$ (resp. $Hom(-,\mathcal{F})$ and $\omega_{\mathbb{P}^1}= \mathcal{O}_{\mathbb{P}^1}(-2)$ the dualizing sheaf of $\mathbb{P}^1$.
My question is how the bilinear map $Hom(\mathcal{O}_{\mathbb{P}^1}, \mathcal{F}) \times Ext^1(\mathcal{F}, \omega_{\mathbb{P}^1}) \to Ext^1(\mathcal{O}_{\mathbb{P}^1}),\omega_{\mathbb{P}^1})$ is explicitely realized?
Let $f\in\mathrm{Hom}(\mathcal{O}_{\mathbb{P}^1},\mathcal{F})$ and $\eta\in \mathrm{Ext}^1(\mathcal{F},\omega_{\mathbb{P}^1})$. So, $\eta$ corresponds to an extension $0\to\omega_{\mathbb{P}^1}\to *\to \mathcal{F}\to 0$, then using $f:\mathcal{O}_{\mathbb{P}^1}\to\mathcal{F}$ taking pull back, you get an extension $0\to \omega_{\mathbb{P}^1}\to *\to\mathcal{O}_{\mathbb{P}^1}\to 0$, which gives an element $\eta'\in\mathrm{Ext}^1(\mathcal{O}_{\mathbb{P}^1},\omega_{\mathbb{P}^1})$. Thus we get the pairing $(f,\eta)\mapsto \eta'$.