I am trying to prove that the functor \begin{align*} \mathrm{Top} &\longrightarrow \mathrm{Cond}(\mathrm{Set}) \\ X &\longmapsto \underline{X} \end{align*} admits a left adjoint and it is the functor $T \mapsto T(*)$ where $T(*)$ has the quotient topology of the map $\bigsqcup_{\underline{S} \rightarrow T} S \rightarrow T(*)$ where the disjoint union runs over all profinite sets $S$ with a map to $T$.

To prove this I am trying to construct the following bijection. We have the map \begin{align*} \phi: \mathrm{Hom}(T, \underline{X}) &\longrightarrow \mathrm{Hom}(T(*), X) \\ f &\longmapsto f_* \ . \end{align*} On the other hand, if $g \in \mathrm{Hom}(T(*), X)$, then we can consider the map $\underline{g}: \underline{T(*)} \rightarrow \underline{X}$. I am trying to construct a map $i: T \rightarrow \underline{T(*)}$ to have an inverse for $\phi$: \begin{align*} \psi: \mathrm{Hom}(T(*), X) &\longrightarrow \mathrm{Hom}(T, \underline{X}) \\ g &\longmapsto \underline{g} \circ i \ . \end{align*} Is there a way to construct $i$? How to prove this adjunction?

Edit: I have managed to construct $i$, but I still could not prove that $\phi$ and $\psi$ are mutually inverses.

I have constructed $i$ as the following: For each $S$ profinite we have the equivalence $\mathrm{Hom}(\underline{S},T) = T(S)$ thanks to the Yoneda Lemma. We can define

\begin{align*} i_S: \mathrm{Hom}(\underline{S},T) &\longrightarrow \mathrm{Hom}(S, T(*)) \\ \eta &\longmapsto \eta_* \ . \end{align*}

It is easy to prove that for $g \in \mathrm{Hom}(T(*), X)$ we have $(\underline{g} \circ i)_* = g$, but have not managed to prove that for $f \in \mathrm{Hom}(T, \underline{X})$ we have $f = \underline{f_*} \circ i$. So it only remains to prove that for $\eta \in \mathrm{Hom}(\underline{S},T)$ we have

\begin{align*} f_S (\eta) = f_* \circ \eta_* \end{align*}

I suspect this is due to the Yoneda Lemma but I could not prove it.