Adjunction between topological spaces and condensed sets

Question

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.

Answer 1

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.