Covering of genus 2 of elliptic curve I am reading this paper and I get confused somewhere:
https://link.springer.com/chapter/10.1007%2F978-1-4612-0457-2_7
So the set up is the following:
Let $C$ be a genus 2 curve. Suppose $C$ admits a degree $n$ covering of elliptic curve $E_{1}$ over a field $K$:
$$\phi_{1}: C \to E_{1}$$
where $n$ prime to char $K$. Assume there is no non tirivial unramified covering between $E_{1}$ and $C$. Let $J$ be the jacobian of $C$. Then there are induced maps by $\phi_{1}$:
$$\phi_{1}^{*}:E_{1} \to J$$
$$\phi_{1,*}:J \to E_{1}$$
(The authors didn't say how these maps work explicitly, my understanding of these two maps are: $\phi_{1}^{*}$ is the pullback from $E_{1}$ to $C^{n}$ and then map to $J$; $\phi_{1,*}$ maps $J$ to $C^{n}$ then each factor of $C^{n}$ maps to $E_{1}$ and add up in $E_{1}$.)
The author said that the maximality of $C \to E_{1}$ implies that $\phi_{1}$ is injective and that $ker(\phi_{1,*})$ is an elliptic curve $E_{2}^{*}$ (cf. Serre). Why this $\phi_{1}$ is injective? Isn't it a degree $n$ covering so one should expect $n$ preimages generically?
The projection map $\phi_{2,*}: J \to J/E_{1}^{*}=:E_{2}$ induces a covering
$$\phi_{2}:C \to E_{2}$$ 
of degree $n$. And the map 
$$h:\phi_{1}^{*} \times \phi_{2}^{*}:E_{1} \times E_{2} \to J$$
is an isogeny of degree $n^{2}$. Why it is of degree $n^{2}$? I've tried the sum of pullbacks but it looks not easy to find (explitcitly) the $n^{2}$ kernel. Anyone knows how this map works on elements?
I very appreciate your help!
 A: First, about the induced maps by $\phi_{1}$:
$$\phi_{1}^{*}:E_{1} \to J$$
$$\phi_{1,*}:J \to E_{1}$$
the correspond, $\phi_1^*$ to universal property of the jacobian: to a map $f:C\to C'$ of curves one gets a map $f^*:J(C')\to J(C)$ of the jacobians. If you identify the points of $J(C)$ with the degree 0 divisors $\text{Div}^0(C)$, then the map sends a point $P$ to the preimatge divisor $f^*(P)=\sum_{f(Q)=P} Q$ (with the multiplicities) and extend it by linearity. The map $\phi_{1,*}$ is the dual map ${\phi_1^*}^{\vee}$, identifying $J(C)^{\vee}$ with $J(C)$ via the principal polarization; it coincides also with the universal map for the Albanese property (so $J(C)$ is also the Albanese variety of $C$). 
That $\phi_1$ is injective is clearly a typo: they meant $\phi_1^*$. 
Now, to show that $h$ is a degree $n^2$ isogeny you can try the following: first, we compute the composition $$\phi_{1,*}\circ \phi_{1}^{*}: E_1\to E_1$$ as follows: given any point $P\in E_1(K)$ (say $K$ a field), $\phi_1^*(P)=\phi_1^{-1}(P)-\phi_1^{-1}(O)$ (being $O$ the zero point of $E_1$), and then 
$$\phi_{1,*}(\phi_1^*(P))=\sum_{\phi_1(Q)=P}\phi_1(Q) - \sum_{\phi_1(Q)=O}\phi_1(Q)=nP$$
(I have been a bit sloppy identifying the points $P$ in $E$ and the degree 0 divisors $P-O$ in $E$). SO we get the $$\phi_{1,*}\circ \phi_{1}^{*}=[n]_{E_1}$$ is the multiplication by $n$ in $E_1$. Same argument applies to $\phi_2$, and so we get that 
$$h^{\vee}\circ h: E_1\times E_2\to E_1\times E_2$$
is multiplication by $n$ in $E_1\times E_2$ (as $h^{\vee}=\phi_{1,*}\times\phi_{2,*}$). This shows $h$ is a degree $n^2$ isogeny. 
