Given an isogeny $f : E \ra E'$ of elliptic curves of degree $n$ the dual isogeny is an isogeny $\hat{f} : E' \ra E$ of the same degree such that $f \circ \hat{f} = [n]$ Here $[n]$ denotes the multiplication-by-$n$ isogeny $e\mapsto ne$ which has degree $n^2$

Often only the existence of a dual isogeny is needed, but the construction is explicit as $$E'\ra \Div^0(E')\stackrel{f^*}{\ra}\Div^0(E)\ra E$$ where $\Div^0$ is the group of divisors of degree 0. To do this, we need maps $E \ra \Div^0(E)$ given by $P\mapsto P - O$ where $O$ is the neutral point of $E$ and $\Div^0(E) \ra E$ given by $\sum n_P P \mapsto \sum n_P P$

To see that $f \circ \hat{f} = [n]$ note that the original isogeny $f$ can be written as a composite $$E \ra \Div^0(E)\stackrel{f_*}{\ra} \Div^0(E')\ra E'$$ and that since $f$ is finite of degree $n$ $f_* f^*$ is multiplication by $n$ on $\Div^0(E')$

Alternatively, we can use the smaller Picard group $\Pic^0$ a quotient of $\Div^0$ The map $E\ra \Div^0(E)$ descends to an isomorphism, $E\stackrel{\sim}{\ra}\Pic^0(E)$ The dual isogeny is $$E' \stackrel{\sim}{\ra} \Pic^0(E')\stackrel{f^*}{\ra}\Pic^0(E)\stackrel{\sim}{\ra} E$$

Note that the relation $f \circ \hat{f} = [n]$ also implies the conjugate relation $\hat{f} \circ f = [n]$ Indeed, let $\phi = \hat{f} \circ f$ Then $\phi \circ \hat{f} = \hat{f} \circ [n] = [n] \circ \hat{f}$ But $\hat{f}$ is surjective, so we must have $\phi = [n]$