Universal property for the indiscrete topology I'm learning about category theory and the universal properties and have the following example.

Given a set $S$ we can build a topological space $D(S)$ by equipping $S$ with the discrete topology: all subsets are open. With this topology,
any map from $S$ to a space $X$ is continuous.

And the universal property given is $$\require{AMScd}
\def\diaguparrow#1{\smash{
  \raise.6em\rlap{\scriptstyle #1}
  \lower.3em{\mathord{\diagup}}
  \raise.52em{\!\mathord{\nearrow}}
}}
\begin{CD}
&& \forall X\\
& \diaguparrow{\forall f} @AA\ \exists! \text{ continuous } \tilde{f} A  \\
S @>> \iota> D(S)
\end{CD}$$
I'm trying to do a similar argument for the indiscrete topology, but my confusion is really about what is being "given" to me and what is not.
If I have a set $S$ and give it the indiscrete topology I get a space say $I(S)$. Then I know that any map to this space is continuous as the only open sets of $I(S)$ are $S$ and $\emptyset.$ So for any $f:Y \to I(S)$ I have that $f^{-1}(S)$ and $f^{-1}(\emptyset)$ are open.
How do I construct the universal property here?
 A: Call $S^\delta$ the set with the discrete topology.
The property that every function $S\to X$ extends to a unique continuous function $S^\delta \to X$ along the map that you call $\iota$, and I will call $\eta$, dualises saying that every function $Y\to S$ lifts to a continuous function $Y\to S^\iota$ ($\iota$=indiscrete), along a canonical map $\epsilon : S^\iota\to S$; this is the sense in which the diagram is "reversed". You will easily draw the diagram witnessing this second property.
This universal property is reversed: come back to this question when you are familiar with adjoint functors, and observe that

*

*$\eta$ is a unit map, and the universal property you write says that $(-)^\delta$ is a left adjoint to the forgetful functor $U : {\sf Top}\to {\sf Set}$ sending a space to its set of points.


*$\epsilon$ is a counit map, and the universal property characterises $(-)^\iota$ as a right adjoint to $U$.
In case you're already familiar with adjunctions, don't go away and prove point 1 and 2!
