= Solution
The <left adjoint> is discretization $D$: give the underlying set of $X$ the <discrete topology>. Write $J$ for indiscretization. Every function $DX\to Y$ is continuous, and every function $X\to JY$ is continuous. The common set of underlying functions supplies the natural bijection
$$
\mathbf{Top}(DX,Y)\cong\mathbf{Top}(X,JY).
$$
The <discrete-indiscrete adjunction> has induced <monad> $JD$, which assigns the indiscrete topology to the original set. Applying it twice gives the same space, and its multiplication is the identity on that set. Thus \b[the adjunction is idempotent]. Its dual comonad $DJ$, discretization, is idempotent as well.
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