Directed Hamiltonian cycle polynomial-time many-one reduces to undirected Hamiltonian cycle

If $\mathrm{\mid }V\mathrm{\mid }\le 1$, decide directly whether $D$ has a directed Hamiltonian cycle and map it to a fixed yes-instance or no-instance of undirected Hamiltonian cycle accordingly. Thus it is enough to treat the case $\mathrm{\mid }V\mathrm{\mid }\ge 2$. Delete every loop, since a Hamiltonian cycle through at least two distinct vertices cannot use a loop. For each vertex $v\in V$, create three undirected vertices $v_1,v_2,v_3$ and the two edges $v_1{v}_2$ and $v_2{v}_3$. For each arc $(u,v)\in A$ with $u\mathrm{\ne }v$, add the undirected edge $u_3{v}_1$. Call the resulting graph $G_D$.

In any Hamiltonian cycle of $G_D$, the middle vertex $v_2$ has degree two, so the cycle must traverse the gadget for $v$ as the consecutive path $v_1,v_2,v_3$ or as the consecutive path $v_3,v_2,v_1$. Moreover, every external gadget edge joins a $3$-vertex to a $1$-vertex. Hence, after orienting the cycle, once one gadget is traversed in one of those two directions, every subsequent gadget is forced to be traversed in the same direction around the cycle.

If $D$ has a directed Hamiltonian cycle $v^{(1)}\to v^{(2)}\to \cdots \to v^{(n)}\to v^{(1)}$, then $v_1^{(1)},v_2^{(1)},v_3^{(1)},v_1^{(2)},v_2^{(2)},v_3^{(2)},\dots ,v_1^{(n)},v_2^{(n)},v_3^{(n)},v_1^{(1)}$ is a Hamiltonian cycle in $G_D$, because each directed arc $v^{(i)}\to v^{(i+1)}$ supplies the edge $v_3^{(i)}{v}_1^{(i+1)}$.

Conversely, let $C$ be a Hamiltonian cycle in $G_D$. By step 2.1 and, if necessary, reversing the orientation of $C$, we may assume that every gadget is traversed as $v_1,v_2,v_3$. Whenever $C$ moves from the gadget of $u$ to the gadget of $v$, it uses an edge $u_3{v}_1$, and such an edge exists only when $(u,v)\in A$. Replacing each gadget path $v_1,v_2,v_3$ by the single vertex $v$ therefore projects $C$ to a directed Hamiltonian cycle of $D$.

The map $D\mapsto G_D$ uses $3\mathrm{\mid }V\mathrm{\mid }$ vertices and at most $\mathrm{\mid }A\mathrm{\mid }+2\mathrm{\mid }V\mathrm{\mid }$ edges, so it is polynomial-time. Steps 2.2 and 3.1 prove $D$ has a directed Hamiltonian cycle if and only if $G_D$ has an undirected Hamiltonian cycle. By [L1], this is a polynomial-time many-one reduction.