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.



Adjunction between topological spaces and condensed sets
As Wojowu noted in the comments, one should really look at $T\_1$ topological spaces. Consider the functors
\begin{align\*}
G\!: \mathbf{Top}\_{T\_1} &\leftrightarrows \mathbf{Cond}\_\kappa:\!F\\
X &\mapsto \big(\underline X \colon S \mapsto \operatorname{Cont}(S,X)\big)\\
T(\*) &\leftarrow\!\shortmid T,
\end{align\*}
where $T(\*)$ is topologised with the quotient topology via the surjection
$$\pi \colon \coprod\_{(S,f \in T(S))} S \to T(\*)\tag{1}\label{1}$$
given on the component $f \in T(S)$ by $f \colon S \to T(\*)$. By the Yoneda lemma, this really means that $f \colon h\_S \to T$ is a morphism from the representable sheaf $h\_S$ to $T$, and the map $S \to T(\*)$ is the set-theoretic map $f\_{\{\*\}} \colon h\_S(\*) \to T(\*)$. But there is the consistent abuse of notation to denote $h\_S = \underline S$ as $S$. To see that (\ref{1}) is surjective, use $S = \{\*\}$.


As Dylan Wilson noted in the comments, the unit $\eta \colon 1 \to GF$ is given by the natural transformation
\begin{align\*}
(\eta\_T)\_S \colon T(S) &\to \operatorname{Cont}(S,T(\*)) \\
f &\mapsto \pi \circ \iota\_f,
\end{align\*}
where $\iota\_f \colon S \to \displaystyle\coprod\_{(S',f')} S'$ is the insertion of the coordinate corresponding to $(S,f)$. This gives the maps
\begin{align\*}
\phi\!: \operatorname{Hom}\_{\mathbf{Cond}}(T,\underline X) &\leftrightarrows \operatorname{Cont}(T(\*),X) :\!\psi \\
f &\mapsto f\_{\{\*\}} \\
g \circ \eta\_T &\leftarrow\!\shortmid g.
\end{align\*}
It remains to check that these are inverses. If $g \colon T(\*) \to X$ is continuous, then $\phi(\psi(g)) \colon T(\*) \to X$ is given by $\psi(g)\_{\{\*\}}$, i.e. the map taking $t \in T(\*)$ to $g \circ \pi \circ \iota\_f \colon \{\*\} \to X$. But $\pi \circ g \colon \{\*\} \to T(\*)$ is just (the constant map with value) the point $t$ (this is how we checked surjectivity of $\pi$ earlier!), so $\psi(g)\_{\{\*\}}$ takes $t$ to $g(t)$, i.e. $\phi(\psi(g)) = g$.


Conversely, given a natural transformation $f \colon T \to \underline X$, we get another natural transformation $\psi(\phi(f)) \colon T \to \underline X$. Write $g \colon T(\*) \to X$ for $\phi(f) = f\_{\{\*\}}$. If $S$ is extremally disconnected and $h \in T(S)$, then $\psi(g)\_S$ takes $h$ to the composition
$$S \overset{\iota\_h}\to \coprod\_{(S',h')} S' \overset\pi\to T(\*) \overset g\to X.$$
By definition, the composition $\pi \circ \iota\_h \colon S \to T(\*)$ is the map $h\_{\{\*\}} \colon h\_S(\*) \to T(\*)$. Thus $\psi(g)\_S(h)$ is the composition
$$h\_S(\*) \overset{h\_{\{\*\}}}\longrightarrow T(\*) \overset{f\_{\{\*\}}}\longrightarrow X.$$
This is the same thing as $(fh)\_{\{\*\}} \colon h\_S(\*) \to \underline X(\*)$, which is the continuous map $S \to X$ given by $f\_S(h) \in \underline X(S)$. We conclude that $\psi(\phi(f))\_S(h) = f\_S(h)$, and since $S$ and $h$ are arbitrary that $\psi(\phi(f)) = f$. $\square$


**Remark.** Morally what's going on here is the following: since $T(\*)$ has the quotient topology, a map $g \colon T(\*) \to X$ is continuous if and only if each of the compositions $gf \colon S \to X$ are continuous with $S$ extremally disconnected and $f \colon S \to T$ an $S$-point of $T$. Thus, a continuous map $T(\*) \to X$ is the same as continuous maps $S \to X$ for every $S \to T$, which is roughly what a natural transformation $T \to \underline X$ is.


