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Now the question is to express the modular units $X$ and $Y$ in terms of $F$ and $G$. Since Yang's construction uses explicit modular units, we know the $q$-expansions of $X,Y$ and $F,G$. Every rational function on $X_1(N)$ is of the form $\sum_{j=0}^{m-1} R_j(F) G^j$, where the $R_j$ are rational functions. If we know that the $R_j$ have degree $\leq d$, then we can simply use the $q$-expansions to find a linear relation of the form \begin{equation*} (*) \qquad \sum_{i=0}^d a_i F^i X = \sum_{i=0}^d \sum_{j=0}^{m-1} b_{i,j} F^i G^j. \end{equation*} with $a_i, b_{i,j} \in \mathbb{Q}$, and deduce $X$ in terms of $F$ and $G$ (and proceeding similarly for $Y$).

Once the computer detects a relation $A(F,G)X+B(F,G)=0$ as above, then we should certify it. To this end, it suffices to show that the function $R=A(F,G)X-B(F,G)$ is regular at all cusps of $X_1(N)$ above $\infty \in X_0(N)$, and is zero at $\infty \in X_1(N)$. Indeed, the function $R$ is already regular away from these cusps. The cusps of $X_1(N)$ above $\infty \in X_0(N)$ are acted on simply transitively by the diamond automorphisms $\langle \gamma \rangle : X_1(N) \to X_1(N)$ with $\gamma \in \Gamma_0(N)/\pm \Gamma_1(N)$, the last group being isomorphic to $(\mathbb{Z}/N\mathbb{Z})^\times / \pm 1$. Checking regularity at these cusps can be done using the transformation formulas for modular units (Proposition 2 in Yang's article). An alternative way is to bound the orders of the poles of $X$ at these cusps (these orders are equal and we already know them since $X$ comes from $X_0(N)$). Then we just need to check that the order of vanishing of $R$ at $\infty \in X_1(N)$ (which can be done by computing the $q$-expansion to enough accuracy) is greater than the sum of the orders of the other possible poles.

It would be nice to give an a priori bound on the degrees of the $R_j$ (that is, on $d$). It is not difficult to find a polynomial $H$ in $F$ such that $HX$ is regular away from infinity. The idea is to cancel the poles of $X$ at the cusps $\langle \gamma \rangle \infty \in X_1(N) \backslash \{\infty\}$, by multiplying by suitable powers of $F-F(\langle \gamma \rangle \infty)$. Yet is not clear to me that the resulting function $HX$ is a polynomial in $F,G$, because the curve $P(F,G)=0$ may be singular, so that its ring of regular functions may be a strict subring in the normalisation $\mathcal{O}(X_1(N) \backslash \infty)$. However, searching for a relation as in $(*)$ should work in practice, by increasing the value of $d$ progressively if necessary. As a variant of the above procedure, we may also search for a relation of the form $A(F,G)X=B(F,G)$.

What this answer doesn't address is whether the pairs $(F,G)$ and $(X,Y)$ can be chosen consistently, that is, in such a way that the resulting equation for $X_1(N) \to X_0(N)$ is as simple as possible. In fact, once $(F,G)$ has been found, there is a standard choice, namely $X=\sum_\gamma F | \gamma$ and $Y=\sum_\gamma G | \gamma$, where the average is over the diamond automorphisms. These are modular funtions on $X_0(N)$ and as Yang explains, the functions $X$ and $Y$ satisfy the assumptions of his theorem, so they generate the function field of $X_0(N)$ and we can compute the resulting model of $X_0(N)$. Then the method above gives an equation for $X_1(N) \to X_0(N)$. I don't know whether this choice of $(X,Y)$ produces simpler equations.

With the same technique, we can also express $F|\gamma$ and $G|\gamma$ in terms of $F$ and $G$ (recall that the $q$-expansions of $F|\gamma$ and $G|\gamma$ can be obtained using the transformation formulas for modular units). This gives, as a bonus, equations for the diamond automorphisms $\langle \gamma \rangle : X_1(N) \to X_1(N)$.

Now the question is to express the modular units $X$ and $Y$ in terms of $F$ and $G$. Since Yang's construction uses explicit modular units, we know the $q$-expansions of $X,Y$ and $F,G$. Every rational function on $X_1(N)$ is of the form $\sum_{j=0}^{m-1} R_j(F) G^j$, where the $R_j$ are rational functions. In the case of $X$, if we know that the $R_j$ have degree $\leq d$, then we can simply use the $q$-expansions to find a linear relation of the form \begin{equation*} (*) \qquad \sum_{i=0}^d a_i F^i X = \sum_{i=0}^d \sum_{j=0}^{m-1} b_{i,j} F^i G^j. \end{equation*} with $a_i, b_{i,j} \in \mathbb{Q}$, and deduce $X$ in terms of $F$ and $G$ (and proceeding similarly for $Y$). As a variant, we may also search for a relation of the form $A(F,G)X=B(F,G)$.

Once the computer detects a relation $A(F,G)X=B(F,G)$ as above, then we should certify it. To this end, it suffices to show that the function $R=A(F,G)X-B(F,G)$ is regular at all cusps of $X_1(N)$ above $\infty \in X_0(N)$, and is zero at $\infty \in X_1(N)$. Indeed, the function $R$ is already regular away from these cusps. The cusps of $X_1(N)$ above $\infty \in X_0(N)$ are acted on simply transitively by the diamond automorphisms $\langle \gamma \rangle : X_1(N) \to X_1(N)$ with $\gamma \in \Gamma_0(N)/\pm \Gamma_1(N)$, the last group being isomorphic to $(\mathbb{Z}/N\mathbb{Z})^\times / \pm 1$. Checking regularity at these cusps can be done using the transformation formulas for modular units (Proposition 2 in Yang's article). An alternative way is to bound the orders of the poles of $X$ at these cusps. In fact these orders are equal since $X$ comes from $X_0(N)$, so we already know them. Then we just need to check that the order of vanishing of $R$ at $\infty \in X_1(N)$ is greater than the sum of the orders of the other possible poles. This can be done by computing the $q$-expansion to enough accuracy.

It would be nice to give an a priori bound on the degrees of the $R_j$ (that is, a bound on $d$). It is not difficult to find a polynomial $H$ in $F$ such that $HX$ is regular away from infinity. The idea is to cancel the poles of $X$ at the cusps $\langle \gamma \rangle \infty \in X_1(N) \backslash \{\infty\}$, by multiplying by suitable powers of $F-F(\langle \gamma \rangle \infty)$. Yet is not clear to me that the resulting function $HX$ is a polynomial in $F,G$, because the curve $P(F,G)=0$ may be singular, so that its ring of regular functions may be a strict subring in the normalisation $\mathcal{O}(X_1(N) \backslash \infty)$. However, searching for a relation as in $(*)$ should work in practice, by increasing the value of $d$ progressively if necessary.

What this answer doesn't address is whether the pairs $(F,G)$ and $(X,Y)$ can be chosen consistently, that is, in such a way that the resulting equation for $X_1(N) \to X_0(N)$ is as simple as possible. In fact, once $(F,G)$ has been found, there is a standard choice, namely $X=\sum_\gamma F | \gamma$ and $Y=\sum_\gamma G | \gamma$, where the average is over the diamond automorphisms. These are modular funtions on $X_0(N)$ and as Yang explains, the functions $X$ and $Y$ satisfy the assumptions of his theorem, so they generate the function field of $X_0(N)$ and we can compute the resulting model of $X_0(N)$. Then the method above gives an equation for the morphism $X_1(N) \to X_0(N)$. I don't know whether this choice of $(X,Y)$ produces simpler equations.

With the same technique, we can also express $F|\gamma$ and $G|\gamma$ in terms of $F$ and $G$ (recall that the $q$-expansions of $F|\gamma$ and $G|\gamma$ can be obtained using the transformation formulas for modular units). As a bonus, this gives equations for the diamond automorphisms $\langle \gamma \rangle : X_1(N) \to X_1(N)$. Note that if $(\mathbb{Z}/N\mathbb{Z})^\times / \pm 1$ is cyclic, generated by $\gamma_0$, then we only need to compute $F|\gamma_0$ and $G|\gamma_0$ in terms of $F$ and $G$.

It is possible to find equations for $X_1(N) \to X_0(N)$ using work of Yifan Yang. In the article Defining equations of modular curves, he explains an algorithm to obtain equations for the modular curves $X_1(N)$ and $X_0(N)$. Let me explain how to use his results to get an equation for $\pi : X_1(N) \to X_0(N)$.

Now the question is to express the modular units $X$ and $Y$ in terms of $F$ and $G$. Since Yang's construction uses explicit modular units, we know the $q$-expansions of $X,Y$ and $F,G$. Every rational function on $X_1(N)$ is of the form $\sum_{j=0}^{m-1} R_j(F) G^j$, where the $R_i$ are rational functions. If we know that the $R_i$ have degree $\leq d$, then we can simply use the $q$-expansions to find a linear relation of the form \begin{equation*} \sum_{i=0}^d F^i X = \sum_{i=0}^d \sum_{j=0}^{m-1} F^i G^j. \end{equation*} and deduce $X$ in terms of $F$ and $G$ (and similarly for $Y$). It would be nice to give an a priori bound on the degrees of the $R_j$. It is not difficult to find a polynomial $H$ in $F$ such that $HX$ is regular away from infinity. The idea is to cancel the poles of $X$ at the cusps of $X_1(N)$ lying above $\infty \in X_0(N)$ but not equal to $\infty$, by multiplying by suitable powers of $F-F(\gamma \infty)$ with $\gamma \in \Gamma_0(N)$. Yet is not clear to me that the resulting function $HX$ is a polynomial in $F,G$, because the curve $P(F,G)=0$ may be singular, so that its ring of regular functions may be a strict subring of $\mathcal{O}(Y_1(N) \backslash \infty)$. However, searching for a relation as above should work in practice, by increasing the value of $d$ progressively if necessary.

I should add that once the computer detects a relation $A(F)X+B(F,G)=0$ as above, then to certify it, it suffices to show that the function $A(F)X+B(F,G)$ is regular at all cusps $\gamma \infty$ with $\gamma \in \Gamma_0(N)$, and is zero at $\infty$. Indeed, this function is already regular away from these cusps. Checking regularity can be done using the transformation formulas for modular units (Proposition 2 in the article). An alternative way is to bound the orders of the poles of $X$ at the cusps $\gamma \infty$ with $\gamma \in \Gamma_0(N)$ (these orders are the same since $X$ comes from $X_0(N)$, and we already know the order of the pole at $\infty$). Then we just need to check that the order of vanishing of $A(F)X+B(F,G)$ is greater than the sum of the orders of these possible poles.

I haven't tried this method, but since Yang's functions and equations are pretty simple, we may hope the same for the equation of $X_1(N) \to X_0(N)$.

It is possible to find equations for $\pi : X_1(N) \to X_0(N)$ using work of Yifan Yang. In the article Defining equations of modular curves, he explains an algorithm to obtain equations for the modular curves $X_1(N)$ and $X_0(N)$. Let me explain how to use his results to get an equation for $\pi$.

Now the question is to express the modular units $X$ and $Y$ in terms of $F$ and $G$. Since Yang's construction uses explicit modular units, we know the $q$-expansions of $X,Y$ and $F,G$. Every rational function on $X_1(N)$ is of the form $\sum_{j=0}^{m-1} R_j(F) G^j$, where the $R_j$ are rational functions. If we know that the $R_j$ have degree $\leq d$, then we can simply use the $q$-expansions to find a linear relation of the form \begin{equation*} (*) \qquad \sum_{i=0}^d a_i F^i X = \sum_{i=0}^d \sum_{j=0}^{m-1} b_{i,j} F^i G^j. \end{equation*} with $a_i, b_{i,j} \in \mathbb{Q}$, and deduce $X$ in terms of $F$ and $G$ (and proceeding similarly for $Y$).

Once the computer detects a relation $A(F,G)X+B(F,G)=0$ as above, then we should certify it. To this end, it suffices to show that the function $R=A(F,G)X-B(F,G)$ is regular at all cusps of $X_1(N)$ above $\infty \in X_0(N)$, and is zero at $\infty \in X_1(N)$. Indeed, the function $R$ is already regular away from these cusps. The cusps of $X_1(N)$ above $\infty \in X_0(N)$ are acted on simply transitively by the diamond automorphisms $\langle \gamma \rangle : X_1(N) \to X_1(N)$ with $\gamma \in \Gamma_0(N)/\pm \Gamma_1(N)$, the last group being isomorphic to $(\mathbb{Z}/N\mathbb{Z})^\times / \pm 1$. Checking regularity at these cusps can be done using the transformation formulas for modular units (Proposition 2 in Yang's article). An alternative way is to bound the orders of the poles of $X$ at these cusps (these orders are equal and we already know them since $X$ comes from $X_0(N)$). Then we just need to check that the order of vanishing of $R$ at $\infty \in X_1(N)$ (which can be done by computing the $q$-expansion to enough accuracy) is greater than the sum of the orders of the other possible poles.

It would be nice to give an a priori bound on the degrees of the $R_j$ (that is, on $d$). It is not difficult to find a polynomial $H$ in $F$ such that $HX$ is regular away from infinity. The idea is to cancel the poles of $X$ at the cusps $\langle \gamma \rangle \infty \in X_1(N) \backslash \{\infty\}$, by multiplying by suitable powers of $F-F(\langle \gamma \rangle \infty)$. Yet is not clear to me that the resulting function $HX$ is a polynomial in $F,G$, because the curve $P(F,G)=0$ may be singular, so that its ring of regular functions may be a strict subring in the normalisation $\mathcal{O}(X_1(N) \backslash \infty)$. However, searching for a relation as in $(*)$ should work in practice, by increasing the value of $d$ progressively if necessary. As a variant of the above procedure, we may also search for a relation of the form $A(F,G)X=B(F,G)$.

What this answer doesn't address is whether the pairs $(F,G)$ and $(X,Y)$ can be chosen consistently, that is, in such a way that the resulting equation for $X_1(N) \to X_0(N)$ is as simple as possible. In fact, once $(F,G)$ has been found, there is a standard choice, namely $X=\sum_\gamma F | \gamma$ and $Y=\sum_\gamma G | \gamma$, where the average is over the diamond automorphisms. These are modular funtions on $X_0(N)$ and as Yang explains, the functions $X$ and $Y$ satisfy the assumptions of his theorem, so they generate the function field of $X_0(N)$ and we can compute the resulting model of $X_0(N)$. Then the method above gives an equation for $X_1(N) \to X_0(N)$. I don't know whether this choice of $(X,Y)$ produces simpler equations.

With the same technique, we can also express $F|\gamma$ and $G|\gamma$ in terms of $F$ and $G$ (recall that the $q$-expansions of $F|\gamma$ and $G|\gamma$ can be obtained using the transformation formulas for modular units). This gives, as a bonus, equations for the diamond automorphisms $\langle \gamma \rangle : X_1(N) \to X_1(N)$.

I haven't done experiments with this method, but since Yang's functions and equations are pretty simple, we may hope the same for the equation of the morphism $X_1(N) \to X_0(N)$.

It is possible to find equations for $X_1(N) \to X_0(N)$ using work of Yifan Yang. In the article Defining equations of modular curves, he explains an algorithm to obtain equations for the modular curves $X_1(N)$ and $X_0(N)$. Let me explain how to use his results to get an equation for $\pi : X_1(N) \to X_0(N)$.

Yang shows that there exist two modular units $F,G$ on $X_1(N)$ such that $F$ and $G$ are regular away from the cusp at infinity, and the orders of the poles of $F$ and $G$ at this cusp are relatively prime. Then $F$ and $G$ generate the function field of $X_1(N)$ and using the $q$-expansions of $F$ and $G$, it is easy linear algebra to compute an equation $P(F,G)=0$ of $X_1(N)$. If $F$ has a pole of order $m$ and $G$ has a pole of order $n$, then the equation has degree $n$ in $F$, and $m$ in $G$. The same can be done for $X_0(N)$, giving an equation $Q(X,Y)=0$.

Now the question is to express the modular units $X$ and $Y$ in terms of $F$ and $G$. Since Yang's construction uses explicit modular units, we know the $q$-expansions of $X,Y$ and $F,G$. Every rational function on $X_1(N)$ is of the form $\sum_{j=0}^{m-1} R_j(F) G^j$, where the $R_i$ are rational functions. If we know that the $R_i$ have degree $\leq d$, then we can simply use the $q$-expansions to find a linear relation of the form \begin{equation*} \sum_{i=0}^d F^i X = \sum_{i=0}^d \sum_{j=0}^{m-1} F^i G^j. \end{equation*} and deduce $X$ in terms of $F$ and $G$ (and similarly for $Y$). It would be nice to give an a priori bound on the degrees of the $R_j$. It is not difficult to find a polynomial $H$ in $F$ such that $HX$ is regular away from infinity. The idea is to cancel the poles of $X$ at the cusps of $X_1(N)$ lying above $\infty \in X_0(N)$ but not equal to $\infty$, by multiplying by suitable powers of $F-F(\gamma \infty)$ with $\gamma \in \Gamma_0(N)$. Yet is not clear to me that the resulting function $HX$ is a polynomial in $F,G$, because the curve $P(F,G)=0$ may be singular, so that its ring of regular functions may be a strict subring of $\mathcal{O}(Y_1(N) \backslash \infty)$. However, searching for a relation as above should work in practice, by increasing the value of $d$ progressively if necessary.

I should add that once the computer detects a relation $A(F)X+B(F,G)=0$ as above, then to certify it, it suffices to show that the function $A(F)X+B(F,G)$ is regular at all cusps $\gamma \infty$ with $\gamma \in \Gamma_0(N)$, and is zero at $\infty$. Indeed, this function is already regular away from these cusps. Checking regularity can be done using the transformation formulas for modular units (Proposition 2 in the article).

I haven't tried this method, but since Yang's functions and equations are pretty simple, we may hope the same for the equation of $X_1(N) \to X_0(N)$.

It is possible to find equations for $X_1(N) \to X_0(N)$ using work of Yifan Yang. In the article Defining equations of modular curves, he explains an algorithm to obtain equations for the modular curves $X_1(N)$ and $X_0(N)$. Let me explain how to use his results to get an equation for $\pi : X_1(N) \to X_0(N)$.

Yang shows that there exist two modular units $F,G$ on $X_1(N)$ such that $F$ and $G$ are regular away from the cusp at infinity, and the orders of the poles of $F$ and $G$ at this cusp are relatively prime. Then $F$ and $G$ generate the function field of $X_1(N)$ and using the $q$-expansions of $F$ and $G$, it is easy linear algebra to compute an equation $P(F,G)=0$ of $X_1(N)$. If $F$ has a pole of order $m$ and $G$ has a pole of order $n$, then the equation has degree $n$ in $F$, and $m$ in $G$. The same can be done for $X_0(N)$, giving an equation $Q(X,Y)=0$.

Now the question is to express the modular units $X$ and $Y$ in terms of $F$ and $G$. Since Yang's construction uses explicit modular units, we know the $q$-expansions of $X,Y$ and $F,G$. Every rational function on $X_1(N)$ is of the form $\sum_{j=0}^{m-1} R_j(F) G^j$, where the $R_i$ are rational functions. If we know that the $R_i$ have degree $\leq d$, then we can simply use the $q$-expansions to find a linear relation of the form \begin{equation*} \sum_{i=0}^d F^i X = \sum_{i=0}^d \sum_{j=0}^{m-1} F^i G^j. \end{equation*} and deduce $X$ in terms of $F$ and $G$ (and similarly for $Y$). It would be nice to give an a priori bound on the degrees of the $R_j$. It is not difficult to find a polynomial $H$ in $F$ such that $HX$ is regular away from infinity. The idea is to cancel the poles of $X$ at the cusps of $X_1(N)$ lying above $\infty \in X_0(N)$ but not equal to $\infty$, by multiplying by suitable powers of $F-F(\gamma \infty)$ with $\gamma \in \Gamma_0(N)$. Yet is not clear to me that the resulting function $HX$ is a polynomial in $F,G$, because the curve $P(F,G)=0$ may be singular, so that its ring of regular functions may be a strict subring of $\mathcal{O}(Y_1(N) \backslash \infty)$. However, searching for a relation as above should work in practice, by increasing the value of $d$ progressively if necessary.

I should add that once the computer detects a relation $A(F)X+B(F,G)=0$ as above, then to certify it, it suffices to show that the function $A(F)X+B(F,G)$ is regular at all cusps $\gamma \infty$ with $\gamma \in \Gamma_0(N)$, and is zero at $\infty$. Indeed, this function is already regular away from these cusps. Checking regularity can be done using the transformation formulas for modular units (Proposition 2 in the article). An alternative way is to bound the orders of the poles of $X$ at the cusps $\gamma \infty$ with $\gamma \in \Gamma_0(N)$ (these orders are the same since $X$ comes from $X_0(N)$, and we already know the order of the pole at $\infty$). Then we just need to check that the order of vanishing of $A(F)X+B(F,G)$ is greater than the sum of the orders of these possible poles.

I haven't tried this method, but since Yang's functions and equations are pretty simple, we may hope the same for the equation of $X_1(N) \to X_0(N)$.