Here's a start on the thrilling **Puzzle 18**. \\(f_\ast\\) does **not** always have a left adjoint, and here's an example.

Let \\(X = \\{1,2,3\\}\\) and \\(Y=\\{A,B\\}\\) and define \\(f\\) by \\(f(1) = A, f(2) = A, f(3) = B\\). \\(f_\ast(\\{1\\}) = \\{A\\}\\), so if the left adjoint \\(g_\ast\\) does exist, then by definition of adjoint \\(g_\ast(\\{A\\}) \subseteq \\{1\\}\\). Similarly \\(g_\ast(\\{A\\}) \subseteq \\{2\\}\\). So \\(g_\ast(\\{A\\}) = \emptyset\\). Thus \\(g_\ast(\\{A\\}) \subseteq \\{3\\}\\), so by definition of adjoint \\(\\{A\\} \subseteq f_\ast(\\{3\\}) = \\{B\\}\\), which is a contradiction.

If \\(f\\) is injective then this sort of argument can't work, and in fact it seems "intuitively obvious" that \\(g_\ast = (f^{-1})_\ast\\) will be a successful left adjoint.

Let \\(X = \\{1,2,3\\}\\) and \\(Y=\\{A,B\\}\\) and define \\(f\\) by \\(f(1) = A, f(2) = A, f(3) = B\\). \\(f_\ast(\\{1\\}) = \\{A\\}\\), so if the left adjoint \\(g_\ast\\) does exist, then by definition of adjoint \\(g_\ast(\\{A\\}) \subseteq \\{1\\}\\). Similarly \\(g_\ast(\\{A\\}) \subseteq \\{2\\}\\). So \\(g_\ast(\\{A\\}) = \emptyset\\). Thus \\(g_\ast(\\{A\\}) \subseteq \\{3\\}\\), so by definition of adjoint \\(\\{A\\} \subseteq f_\ast(\\{3\\}) = \\{B\\}\\), which is a contradiction.

If \\(f\\) is injective then this sort of argument can't work, and in fact it seems "intuitively obvious" that \\(g_\ast = (f^{-1})_\ast\\) will be a successful left adjoint.