Proving that the differential on an elliptic curve $E$ given by $\omega=\frac{dx}{y}$ is translation invariant I'm taking a course on elliptic curves and I'm stuck on a line in a proof. 
We're assuming we're in an algebraically closed field $K$ and char($K)\not=2$. We have our elliptic curve
$$E:y^2=(x-e_1)(x-e_2)(x-e_3)$$
where the $e_i$ are distinct.
We've proved that $\omega=\frac{dx}{y}$ is a regular differential on $E$. 
Now for $P \in E$ we define
$\begin{array}{llll}\tau_P:&E &\rightarrow &E \\ & Q & \mapsto & P \bigoplus Q.\end{array}$
Since $\tau_P^*\omega$ is a regular differential on $E$, we have $\tau_P^*\omega=\lambda_P \omega$ for some $\lambda_P \in K^\times$. Then it is claimed that 

The map $$\begin{array}{lll}E &\rightarrow & \mathbb{P}^1 \\ P & \mapsto & \lambda_P\end{array}$$
  is a morphism of algebraic curves.

Now I'm not sure why this is true. If I could show it was a rational map then I'd be done, since rational maps from smooth projective curves are morphisms. 
Well, so long as $P\not=0_E$ and we're not where the formulae for adding points degenerate, we have (say $P=(x_P,y_P)$)
$\tau_P^*\omega=\frac{d(x \circ \tau_P)}{y\circ \tau_P}=\left(\mbox{rational function of }x, y, x_P\mbox{ and }y_p\right) \frac{dx}{y}$
but I don't see why that rational function doesn't actually have any $x$ or $y$ dependence. Because $\omega$ is a $K$-basis for the space of regular differentials we know this rational function is in $K$. Is that it?

The map $P\mapsto (\lambda_P:1)$ is a rational map because it is given by rational functions (the $\lambda_P$'s). As you pointed out in the question, it is indeed a morphism of algebraic curves. Now, since it is a morphism between nonsingular integral projective curves, it is either surjective or constant. But because $(1:0)$ is not hit, it must be constant. And you can compute, for instance, the value $\lambda_O$, which is $1$. (I named $O$ the origin $(0:1:0)\in E$.) This proves, in passing, that $\tau_P^\ast\omega=\omega$ for every $P\in E$.
