Confusion about normal bundles This is an elementary question about some basic confusion I have.
Consider the Segre embedding $S: \mathbb{P}^{1} \times \mathbb{P}^{1} \rightarrow \mathbb{P}^{3}$. Then the image of a line $\mathbb{P}^{1} \times \{p_{0}\}$ is a line $ L \subset \mathbb{P}^{3}$. Denote the quadric $S(\mathbb{P}^{1} \times \mathbb{P}^{1})$ by $Q$.
The normal bundle of a line in $\mathbb{P}^{3}$ is $\mathcal{O}(1) \oplus \mathcal{O}(1)$. However, considering $L \subset Q \subset \mathbb{P}^{3}$, we should have $N_{L/Q}  =  \mathcal{O} \subset N_{L/\mathbb{P}^{3}}$. However, $\mathcal{O}(1) \oplus \mathcal{O}(1)$ cannot have a subbundle isomorphic to $\mathcal{O}$, where is the mistake?

The exact sequence of normal bundles is:
$$0\to \mathcal{O}\xrightarrow{i} \mathcal{O}(1)\oplus \mathcal{O}(1)\to \mathcal{O}(2)\to 0.$$ 
To understand the inclusion $i$, let's twist it by $\mathcal{O}(-1)$:
$$0\to \mathcal{O}(-1)\xrightarrow{i} \mathcal{O}\oplus \mathcal{O}\xrightarrow{\pi} \mathcal{O}(1)\to 0.$$ 
Does that looks more familiar?
Regard $\mathbb P^1$ as parameter space of lines on $\mathbb A^2$, the middle term is the trivial bundle $\mathbb P^1\times\mathbb A^2$, $i$ is the inclusion of universal subbundle, and $\pi$ is surjective to universal quotient bundle.
