question on adjunction in category $\mathbf{Pos}^{\dashv}$ In the category $\mathbf{Pos}^{\dashv}$, I assume that there is an adjunction $f\dashv g$ with $f,g$ monotonous. Diagrammatically,
$$S \mathrel{\mathop{\leftrightarrows}^{\mathrm{f}}_{\mathrm{g}}} T$$
where $S, T$ are objects in this category. Furthermore, following from definitions of this  category,
$$f(a)\leq b\iff a\leq g(b)$$
Why is it that $$id_S\leq g\circ f$$ and $$f\circ g\leq id_T?$$
The question is from the book on category theory by Harold Simmons, Introduction to Category Theory, Cambridge, 2013, exercise 1.3.6. I have a solution set, but i still cannot work it out why this is the case. In the solutions, it was argued that $a\leq (g\circ f)(a)$ and $(f\circ g)(b)\leq b$, from which the claim follows, using also arguments of monotonicity.
 A: If I'm following this correctly. Each poset can be considered a category and functors between posets are monotone maps.
We consider the category $\mathrm{Pos}^{\dashv}$ whose objects are posets and an arrow $A\to B$ is an adjunction $f\dashv g$, where $f :A\to B$ and $g:B\to A$ satisfy the Galois connection condition. (I presume this is to be proved somewhere else, that the adjunction is described in terms of Galois connections.)
Think of the Galois connection in terms of colours.
$$ f(\color{blue}a)\leqslant _B \color{red}b \Leftrightarrow  \color{blue}a\leqslant _A g(\color{red}b)  $$
Since $\leqslant$ is reflexive, from $f(\color{blue}a)\leqslant _B {\color{red}{f(a)}}$ it follows that
$ \color{blue}a \leqslant _A g({\color{red}{f(a)}}). $
Similarly, $g(b) \leqslant g(b)$ implies $f(g(b)) \leqslant b$. Comparison is defined pointwise, so $\mathrm{id}_A \leqslant gf$ and $fg \leqslant \mathrm{id}_B$.

The converse is true as well. That is, from $\mathrm{id}_A \leqslant gf$ and $fg \leqslant \mathrm{id}_B$ the Galois connection follows because
$$ a\leqslant g(b) \Rightarrow f(a) \leqslant fg(b) \leqslant b $$
and similarly
$$ f(a) \leqslant b \Rightarrow a\leqslant gf(a) \leqslant g(b).$$
