$\renewcommand{\J}{\mathrm{Jac}} \renewcommand{\F}{\mathbb{F}}$ I am reading this paper by B. Gross, and there is something I don't understand on p. 945. Here is the context: fix a prime $p \equiv 3 \pmod 4$, and define the (hyper)elliptic curves over $\F_p$ given by the (affine) equations $$X_1 : y^2 = x^p - x,\quad X_2 : y^2 = x^{p+1}-1,\quad E_1 : y^2 = x^3-x,\quad E_2 : y^2=x^4-1.$$

I checked (using Tate isogeny theorem) that there is a non-zero isogeny $\alpha : \J(X_1) \to \J(X_2)$ between the jacobian varieties (actually both are isogenous to $E_1^{(p-1)/2}$ over $\F_p$), and there is a non-zero isogeny $\beta : E_1 \to E_2$.

Question:There is a morphism $f_2 : X_2 \to E_2, (x,y) \mapsto (x^{(p+1)/4}, y)$ which has degree $(p+1)/4$. Then it is claimed that we therefore get a morphism $f_1 : X_1 \to E_1$ of degree $(p+1)/4$, but I don't see why/how.

**Thoughts:**
I know that $f_1$ induces a morphism $\phi_2 : \J(X_2) \to E_2$, we get a morphism $\beta \circ \phi_2 \circ \alpha^{\vee} : \J(X_1) \to E_1$, hence a morphism $f_1 : X_1 \to E_1$, but I believe that it has degree *at least* the degree of $f_2$.
Maybe there is a clever way to compose $\phi_2$ with other isogenies to get the equality of degrees?

In general, given a non-constant morphism $f_2 : X_2 \to E_2$, it might not be possible to get a morphism $f_1 : X_1 \to E_1$ of same degree as $f_2$: just take $X_2 = E_2 = X_1, f_2 = \mathrm{id}$ and $E_1$ an elliptic curve isogenous but not isomorphic to $E_2$. I am probably missing something easy, but I prefer asking for clarifications.