This is from page 150 of Silverman's "The Arithmetic of Elliptic Curves". Any my only questions is:
How you can conclude that $[a]=\phi+\hat{\phi}$?
I tried to use the formula on page 85 which is:
$[<\phi,\psi>]=[\deg(\phi+\psi)]-[\deg(\phi)]-[\deg(\psi)]=\hat{\phi}\circ \psi+\hat{\psi}\circ \phi$.
And i wanted to show that the LHS is equal to $a$ and the RHS is equal to $\phi+\hat{\phi}$ but it did not work.
Can someone help me?
Consider the multiplication by $a$ map on $E$. You have, by Theorem III.6.2, $$ [a] = [1] - \left( ([1] - \phi) \circ ([1] - \hat{\phi}) \right ) + \phi \circ \hat{\phi}. $$ Using that an isogeny is a homomorphism, you can expand it out to obtain $$ [a] = [1] - [1] + \phi + \hat{\phi} - \phi \circ \hat{\phi} + \phi \circ \hat{\phi}, $$ which simplifies to $[a] = \phi + \hat{\phi}$.