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Put $u=x_1-1$, $v=y_1$. Then $v^2=u(u^2+3u+2)$. After using the curve equation, the displayed rational functions becomeSet $X_0=X+2=u+3+2/u$. Direct substitution into $v^2=u(u^2+3u+2)$ givesover ${\mathbb{F}}_5$. Replacing $X_0$ by $X+2$ makes this $Y^2=X^3-X+1$. This verifies the target equation explicitly, and identifies the functions with the two-isogeny formula after translating the source's two-torsion point to zero and then translating the target coordinate.

We must also check the omitted affine points. At $T=(1,0)$, $v$ is a local parameter, and $u=v^2/(x_1(x_1+1))$ has order two, since $x_1(x_1+1)=2$ at $T$. Thus $X$ has a pole of order two and $Y$ one of order three. At $O_1$, the usual pole orders on a smooth cubic in Weierstrass form give poles two and three as well: the correction terms have higher order and do not cancel their leading terms. There are no other poles. A rational map from a smooth curve into a projective curve extends across these points; the indicated pole orders show that both $T$ and $O_1$ map to $O_2$. Hence this is a nonconstant identity-preserving morphism, and part (i) makes it an isogeny of elliptic curves.

The pullback of the pole divisor $2O_2$ of $X$ is $2O_1+2T$. Degrees of divisors multiply by the morphism degree, so $2\mathrm{deg}\phi =4$. We obtainThe degree is prime to five, so the elliptic isogeny is separable. Equivalently $dX/Y=d{x}_1/y_1$ is a nonzero pullback of an invariant differential.

Use the usual hypothesis that the ground field has characteristic different from two, and work over its algebraic closure for the geometric points. This hypothesis is necessary for the smooth elliptic-curve calculation: in characteristic two, the displayed cubic model can be singular even when its three roots are distinct.

Homogenization givesAt $Z=0$ this says $X^3=0$, soTo compute the pole orders on a smooth cubic in Weierstrass form, work in the chart $Y=1$, writing $u=X/Y$ and $v=Z/Y$. The equation is $v={\prod }_j(u-{\lambda }_{j}v)$. Its derivative in $v$ at $(0,0)$ is one on the left after bringing the right side across, so $u$ is a local parameter. The equation gives $v=u^3+O(u^5)$. Consequently $x=u/v$ and $y=1/v$ have poles of orders two and three at $P_{\infty }$.