Let $E:y^2=x^3+ax+b$ be an elliptic curve over $\mathbb{C}$ with the identity $\mathcal{O}=(0:1:0)$.
It is well known that if $P=(x,y)$ (affine coordinate), then $2P=(f(x,y),g(x,y))$ where $$ f(x,y) =\frac{(3x^2+a)^2-8xy^2}{4y^2}, g(x,y) =\frac{4y^2(7x^2+ax-2b)-(3x^2+a)^3}{8y^3}. $$
This defines the isogeny $[2]:E\to E$.
I want to check that this $[2]$ is a certainly rational map between projective curves (i.e. there are rational functions $f_1,f_2,f_3\in \mathbb{C}(E)$ that are not all zero, $[2]=(f_1:f_2:f_3)$).
I first took homogenization $$ f^*(x,y,z) := f(x/z,y/z) =\frac{(3x^2+az^2)^2-8xy^2z}{4y^2z^2}, $$
$$ g^*(x,y,z) := g(x/z,y/z) =\frac{4y^2z(7x^3+axz^2-2bz^3)-(3x^2+az^2)^3}{8y^3z^3}. $$
Then $$\phi:=(f^*:g^*:1)=(2yz((3x^2+az^2)^2-8xy^2z):4y^2z(7x^3+axz^2-2bz^3)-(3x^2+az^2)^3:8y^3z^3)$$ should be $[2]$ since a rational map between non singular curves is a morphism.
But clearly $\phi(\mathcal{O})\neq \mathcal{O}$, this is weird.
Where did I make a mistake?
Recall that a rational map from $X$ to $Y$ is an equivalence class of pairs of dense open subschemes $U\subset X$ and morphisms $\varphi_U:U\to Y$ where two pairs $(U_1,\varphi_{U_1})$ and $(U_2,\varphi_{U_2})$ are equivalent if the maps are the same on the common intersection of the opens. What you've done here is provide a particular representative for the rational map $[2]$ on a certain open set, but this open set does not contain $\mathcal{O}$: the formula evaluates to $[0:0:0]$ when plugging in $\mathcal{O}$. This isn't necessarily an error, but you do have more to do: you need to play around with this formula to get a representative which is defined at $[0:1:0]$.
Here's how to do that. We have the relation $y^2z-axz^2-bz^3=x^3$, or $(y^2-axz-bz^2)z=x^3$ and we'll let $\sigma=(y^2-axz-bz^2)$ and then multiply all the coordinates of the map by $\sigma^2$:
$$ [2yz((3x^2+az^2)^2-8xy^2z)\sigma^2 : 4y^2z(7x^3+axz^2-2bz^3)\sigma^2-(3x^2+az^2)^3\sigma^2 : 8y^3z^3\sigma^2] $$
Now we can use our relation to write $\sigma z = x^3$, which after a fair amount of algebra will give that all our coordinates are divisible by $x^6$. After factoring out an $x^6$, we're left with $$[x(\cdots)+z(\cdots) : 28y^2\sigma-27\sigma^2+x(\cdots) + z(\cdots) : x(\cdots)+z(\cdots) ]$$ which evaluates to $[0:1:0]$ exactly as it should. (Apologies for not writing out all of the algebra, it gets pretty long.)
Secretly the motivation for what we're doing here is that $x$ is a uniformizer in the local ring of $\mathcal{O}$, so we're rewriting stuff in terms of the uniformizer at least to a point where we can divide through by the uniformizer raised to the minimum valuation of one of the coordinates here, which will then produce a non-vanishing coordinate and therefore a formula valid at $\mathcal{O}$.