Let $f= xy^3+y^4-x^2+xy$. Using the following codes in Maple,

f := xy^3+y^4-x^2+xy;

v := Weierstrassform(f, x, y, x0, y0);

I obtain the following result:

\begin{align} & f_0 = {{ x_0}}^{3}+{{ y_0}}^{2}-{ x_0},\\ & v_2(x,y)={\frac {{y}^{3}}{x}},\\ & v_3(x,y)={y}^{3}-x+2y,\\ & v_4(x_0,y_0)={\frac {{ x_0}{ y_0}+{ y_0}}{{{ x_0}}^{2}-2{ x_0}+1}},\\ & v_5(x_0,y_0)=-{\frac {{ y_0}}{{ x_0}-1}}. \end{align}

Here $f_0 = {{ x_0}}^{3}+{{ y_0}}^{2}-{ x_0}=0$ is the normal form of $f=0$. The result gives an isomorphism from the function field $C(x_0)[y_0]/(f_0)$ to $C(x)[y]/(f)$. $v_2$ is the image of $x_0$ under this isomorphism. $v_3$ is the image of $y_0$ under this isomorphism. $v_4$ is the image of $x$ under the inverse isomorphism. $v_5$ is the image of $y$ under the inverse isomorphism.

The elliptic curve $f_0=0$ has an involution given by $(x_0, y_0) \mapsto (x_0, -y_0)$. I want to compute the corresponding involution $\eta$ on $f=0$. I tried to send $(x, y)$ to $( v_4(v_2(x,y), -v_3(x,y)), v_5(v_2(x,y), -v_3(x,y)) )$. That is, \begin{align} \eta(x,y)=\left( {\frac { \left( -{y}^{3}+x-2\,y \right) x \left( {y}^{3}+x \right) }{ \left( -{y}^{3}+x \right) ^{2}}}, {\frac { \left( -{y}^{3}+x-2\,y \right) x}{-{y}^{3}+x}}\right). \end{align} But the map $\eta$ is not an involution: $\eta^2=1$. How to compute the involution on $f(x,y)=0$ which corresponds to the involution $(x_0, y_0) \mapsto (x_0, -y_0)$ on $f_0(x_0, y_0)=0$? Thank you very much.

Let $f= xy^3+y^4-x^2+xy$. Using the following codes in Maple,

f := xy^3+y^4-x^2+xy;

v := Weierstrassform(f, x, y, x0, y0);

I obtain the following result:

\begin{align} & f_0 = {{ x_0}}^{3}+{{ y_0}}^{2}-{ x_0},\\ & v_2(x,y)={\frac {{y}^{3}}{x}},\\ & v_3(x,y)={y}^{3}-x+2y,\\ & v_4(x_0,y_0)={\frac {{ x_0}{ y_0}+{ y_0}}{{{ x_0}}^{2}-2{ x_0}+1}},\\ & v_5(x_0,y_0)=-{\frac {{ y_0}}{{ x_0}-1}}. \end{align}

Here $f_0 = {{ x_0}}^{3}+{{ y_0}}^{2}-{ x_0}=0$ is the normal form of $f=0$. The result gives an isomorphism from the function field $C(x_0)[y_0]/(f_0)$ to $C(x)[y]/(f)$. $v_2$ is the image of $x_0$ under this isomorphism. $v_3$ is the image of $y_0$ under this isomorphism. $v_4$ is the image of $x$ under the inverse isomorphism. $v_5$ is the image of $y$ under the inverse isomorphism.

The elliptic curve $f_0=0$ has an involution given by $(x_0, y_0) \mapsto (x_0, -y_0)$. I want to compute the corresponding involution $\eta$ on $f=0$. I tried to send $(x, y)$ to $( v_4(v_2(x,y), -v_3(x,y)), v_5(v_2(x,y), -v_3(x,y)) )$. That is, \begin{align} \eta(x,y)=\left( {\frac { \left( -{y}^{3}+x-2\,y \right) x \left( {y}^{3}+x \right) }{ \left( -{y}^{3}+x \right) ^{2}}}, {\frac { \left( -{y}^{3}+x-2\,y \right) x}{-{y}^{3}+x}}\right). \end{align} But the map $\eta$ is not an involution: $\eta^2 \neq 1$. How to compute the involution on $f(x,y)=0$ which corresponds to the involution $(x_0, y_0) \mapsto (x_0, -y_0)$ on $f_0(x_0, y_0)=0$? Thank you very much.