Any smooth projective curve $X$ of genus one admits a morphism $f:X\to\mathbb{P}^1$ of degree 2, which by the Riemann-Hurwitz formula is ramified at exactly four points. Given a $4$-tuple $\{\alpha_1,\alpha_2,\alpha_3,\alpha_4\}\subset\mathbb{P}^1$, I would like to give an explicit such morphism ramified exactly at $\alpha_1,...,\alpha_4$. How can I do this?
For instance, for $\{\alpha_1,\alpha_2,\alpha_3,\infty=[1:0]\}\subset\mathbb{P}^1$ we can consider \begin{equation} X=\{[x:y:z]\in\mathbb{P}^2\mid zy^2=(x-\alpha_1z)(x-\alpha_2z)(x-\alpha_3z)\}, \end{equation} which is the closure in $\mathbb{P}^2$ of the affine curve $\{(x,y)\in\mathbb{C}^2\mid y^2=(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)\}$, and the morphism $F:X\to\mathbb{P}^1$ given by \begin{equation} F([x:y:z]) = \left\{\begin{array}{lr} \left[1:0\right] & \text{if } [x:y:z]=[0:1:0]\\ \left[x:z\right] & \text{otherwise} \end{array}\right\}, \end{equation} which looks like $(x,y)\mapsto x$ on the open chart $\mathbb{C}^2$. This construction satisfies all my requirements (in particular, $X$ is smooth).
However, if I take $\{\alpha_1,..,\alpha_4\}\subset\mathbb{P}^1\setminus\{\infty\}$ and I consider the curve \begin{equation} Y=\{[x:y:z]\in\mathbb{P}^2\mid z^2y^2=(x-\alpha_1z)(x-\alpha_2z)(x-\alpha_3z)(x-\alpha_4z)\},\end{equation} then $Y$ has a singular point at $[0:1:0]$, and the corresponding morphism $Y\to\mathbb{P}^1$ is ramified at $\alpha_1,...,\alpha_4$ $\textbf{and}$ $\infty=[1:0]$, so this construction doesn't satisfy my requirements. So what is the correct construction in this case?