Let's consider the Segre Embedding $$\begin{array}{ccccc} \sigma & : & \mathbb{P}^1 \times \mathbb{P}^1 & \longrightarrow & \mathbb{P}^3 \\ & & [x_0 ; x_1] , [y_0 ; y_1] & \longmapsto & [x_0y_0 ; x_0y_1 ; x_1y_0 ; x_1y_1] \end{array}$$ and let's denote $\Sigma$ the image of Segre Embedding.
Let's consider $[z], [z'] \in \Sigma$. I wanted to show that there exist a morphism $\varphi : \mathbb{P}^1 \to \mathbb{P}^3$ such that $$\varphi(\mathbb{P}^1) \subset \Sigma , \quad \varphi([1 ; 0]) = [z] \quad \text{and} \quad \varphi([0 ; 1]) = [z'].$$
My first thought was to consider the morphism $$\begin{array}{ccccc} \varphi & : & \mathbb{P}^1 & \longrightarrow & \mathbb{P}^3 \\ & & [t,s] & \longmapsto & [tz + sz'] \end{array}$$ It obviouly verifies $\varphi([1 ; 0]) = [z]$ and $\varphi([0 ; 1]) = [z']$. But I have some trouble showing that $\varphi([t ; s])$ belongs to $\Sigma$ for $[t ; s] \in \mathbb{P}^1$. Here is how I started:
Since $[z]$ (resp. $[z']$) belongs to $\Sigma$ then there exist $([x_0 ; x_1], [y_0 ; y_1]) \in \mathbb{P}^1 \times \mathbb{P}^1$ such that $\sigma([x_0 ; x_1], [y_0 ; y_1]) = [z]$ (resp. there exist $([x'_0 ; x_1],[y'_0 ; y'_1]) \in \mathbb{P}^1 \times \mathbb{P}^1$ such that $\sigma([x'_0 ; x'_1], [y'_0 ; y'_1]) = [z']$). More precisely, $$[z] = [x_0y_0 ; x_0y_1 ; x_1y_0 ; x_1y_1] \quad \text{and} \quad [z'] = [x'_0y'_0 ; x'_0y'_1 ; x'_1y'_0 ; x'_1y'_1]$$ So $$\varphi([t ; s]) = [tx_0y_0 + sx'_0y'_0 ; tx_0y_1 + sx'_0y'_1 ; tx_1y_0 + sx'_1y'_0 ; tx_1y_1 + sx'_1y'_1]$$ Showing that $\varphi([t ; s]) \in \Sigma$ is equivalent to show that $$ (tx_0y_0 + sx'_0y'_0)(tx_1y_1 + sx'_1y'_1) = (tx_0y_1 + sx'_0y'_1)(tx_1y_0 + sx'_1y'_0)$$ But when I calculated the first part of the equality, I didn't get the second one.
So I wondered if the morphism I took was the good one and if it is the case, what can I use to find this identity.
Thanks in advance. K. Y.