Following Dan Schmidt's post 5, it seems that if \\(f_{\ast}\\) has a left adjoint \\(g_{\ast}\\), then \\(f\\) must be injective.

Otherwise, if \\(f\\) is not injective, there are elements \\(a_1, a_2 \in A, a_1 \neq a_2\\) such that \\(f(a_1) = b = f(a_2)\\).

Then, \\(f_{\ast}\\) sends both \\(\\{a_1\\}\\) and \\(\\{a_2\\}\\) to \\(\\{b\\}\\).

The left adjoint \\(g_{\ast}: PY \to PX\\) must then send \\(\\{b\\}\\) to something at most \\(\\{a_1\\} \wedge \\{a_2\\} = \emptyset\\), which must be \\(\emptyset\\) itself.

But then we would have \\(\emptyset = g_{\ast}(\\{b\\}) \leq \emptyset\\) while \\(\\{b\\} \nleq f_{\ast}(\emptyset) = \emptyset\\), contradicting our hypothesis.

Otherwise, if \\(f\\) is not injective, there are elements \\(a_1, a_2 \in A, a_1 \neq a_2\\) such that \\(f(a_1) = b = f(a_2)\\).

Then, \\(f_{\ast}\\) sends both \\(\\{a_1\\}\\) and \\(\\{a_2\\}\\) to \\(\\{b\\}\\).

The left adjoint \\(g_{\ast}: PY \to PX\\) must then send \\(\\{b\\}\\) to something at most \\(\\{a_1\\} \wedge \\{a_2\\} = \emptyset\\), which must be \\(\emptyset\\) itself.

But then we would have \\(\emptyset = g_{\ast}(\\{b\\}) \leq \emptyset\\) while \\(\\{b\\} \nleq f_{\ast}(\emptyset) = \emptyset\\), contradicting our hypothesis.