Adjunction between ordinal categories $n$ and $n+1$ Confusion
Consider the ordinal categories $\mathbf n$ and $\mathbf{n + 1}$, given as partial orders generated by $[0 \leq 1 \leq \dots \leq n]$ and $[0 \leq 1 \leq \dots \leq (n+1)]$.
Define a functor $d^0: \mathbf n \to \mathbf{n + 1}$ given by $d^0(k) \equiv k + 1$. (That is, the functor $d^0$ injects $\mathbb n$ into $\mathbb{n+1}$, missing the number $0$). As its adjoint, define a functor $s^0: \mathbf n \to \mathbf{n + 1}$ given by $s^0(k) \equiv \begin{cases} 0 & k = n + 1 \\ k & \text{otherwise} \end{cases}$. That is, it faithfully maps $[0, \dots, n]$ back from $\mathbf{n+1}$ into $\mathbf{n}$, and maps $(n+1)$ to $0$, thereby ensuring that $0$ has two preimages. Riehl claims that $s^0$ is left adjoint to $d^0$. I don't believe that these are adjoints.
Here is the text from Emily Riehl's Category theory in context, example 4.1.14:

Counter-example
We show that $s^0, d^0$ are not adjoint by checking that the composition $s^0 \circ d^0$ is not idempotent, as every Galois Connection should give an idempotent closure operator.  Let's pick $n = 3$ for simplicity. then the mapping $s^0 \circ d^0$ looks like follows:
$$
\begin{matrix}
&\mathbf n &\xrightarrow{d^0} &\mathbf {n + 1}  &\xrightarrow{s^0} &\mathbf {n} \\
&- &\mapsto &0 &\mapsto &0 \\
&0 &\mapsto &1 &\mapsto &1 \\
&1 &\mapsto &2 & \mapsto &2\\
&2 &\mapsto &3 &\mapsto &3
\end{matrix}
$$
Clearly, this mapping is not idempotent. For example, we send $1 \mapsto 2 \mapsto 3 $ upon applying $(s^0 \circ d^0)^2(1)$. Hence, this cannot be a galois connection!
Repairing the example
We define $t^0: \mathbf{n+1} \to \mathbf{n}$ as $t^0(k) \equiv \begin{cases} n & k = 0 \\ k - 1 & \text{otherwise} \end{cases}$. That is, we decrement each number in $\mathbf{n + 1}$. We send $0$ to $n$, since it has no number preceding it. In this case, the composition $(t^0 \circ d^0)$ looks like this:
$$
\begin{matrix}
&\mathbf n &\xrightarrow{d^0} &\mathbf {n + 1}  &\xrightarrow{t^0} &\mathbf {n} \\
&0 &\mapsto &1 &\mapsto &0 \\
&1 &\mapsto &2 & \mapsto &1\\
&2 &\mapsto &3 &\mapsto &2
\end{matrix}
$$
$t^0 \circ d^0: \mathbf n \to \mathbf n$ is clearly idempotent [in fact, it is the identity function]. Let us also compute $d^0 \circ t^0: \mathbf{n+1} \to \mathbf{n+1}$:
$$
\begin{matrix}
&\mathbf n+1 &\xrightarrow{t^0} &\mathbf {n}  &\xrightarrow{d^0} &\mathbf {n+1} \\
&0 &\mapsto &2 &\mapsto &3 \\
&1 &\mapsto &0 &\mapsto &1 \\
&2 &\mapsto &1 & \mapsto &2 \\
&3 &\mapsto &2 &\mapsto &3
\end{matrix}
$$
The function $(d^0 \circ t^0)$ inflates $0$ to $3$, and it fixed $1, 2, 3$. Thus, applying it twice will be the same as applying it once, and is hence idempotent. Thus, we have $t^0 \vdash d^0$, as expected.
Questions

*

*Is the proof that the example shown above is incorrect? If not, what am I missing?

*Is the adjunction that I construct correct?

 A: Your table for $s^0$ doesn't match the text. The key phrase is

for which $i \in \mathbf{n}$ is the unique object with two preimages

So two different objects of $\mathbf{n + 1}$ should map to $0$. Recall that functors between posets are monotonic, and since we want $s^0$ to be surjective, that means that both $0$ and $1$ map to $0$ (leaving $s^0(2) = 1$ and $s^0(3) = 2$). You can check that with this corrected definition, $s^0 \circ d^0$ is idempotent (with a similar calculation for your $t^0$).
Edit: rereading the first part of your question, I see that you have another definition for $s^0$ which takes $n + 1$ to $0$. First, note that $n + 1$ isn't an object of $\mathbf{n + 1}$ (see my comment). Second, even if we take $s^0(n) = 0$, then $s^0$ won't be monotonic (unless $n$ is $1$ or smaller).

The only problem with your proposed $t^0$ is that it sends $0$ to $3$. That means that if $t^0(1) < 3 = t^0(0)$, then $t^0$ isn't monotonic. To fix that, you have to make $t^0(0) = 0$, and then it matches the fixed $s^0$.
