Given an isogeny $f:E\rightarrow E^{{\prime}}$ of elliptic curves of degree $n$, the dual isogeny is an isogeny $\hat{f}:E^{{\prime}}\rightarrow E$ 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

where $\Div^{0}$ is the group of divisors of degree 0. To do this, we need maps $E\rightarrow\Div^{0}(E)$ given by $P\mapsto P-O$ where $O$ is the neutral point of $E$ and $\Div^{0}(E)\rightarrow E$ 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

and that since $f$ is finite of degree $n$, $f_{*}f^{*}$ is multiplication by $n$ on $\Div^{0}(E^{{\prime}})$.

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

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 $\phi=[n]$.