The metric spaces and nonexpansive maps: no non-empty space is finitelly presentable I have asked this question elsewhere, but I wrongly formulated it. Let me now correct it.
Why is no nonempty metric space in the category of metric spaces and non-expansive maps
finitely presentable ?
I got this hint:
Let in the category $\operatorname{ Met}$ of the metric spaces and non-expansive maps
be the diagram with a sequence of $2$ elements approaching with the distance $\frac{1}{n}$.
First: why there is a unique point in the colimit ?
Second: where have we used that the maps are non-expansive in the essential uniqueness of the factorization; why this doesn't hold for continuous maps or even arbitrary functions ?
 A: For $n \in \mathbb{N}$ let $X_n = \{a_n, b_n\}$ be the metric space with two points with distance $\frac{1}{n}$. For $n \leq m$ we have the obvious map $f_{nm}: X_n \to X_m$ given by $f_{nm}(a_n) = a_m$ and $f_{nm}(b_n) = b_m$. With these maps $(X_n)_{n \in \mathbb{N}}$ forms a directed diagram. Let $X = \{*\}$ be the one-point space, then it clearly forms a cocone of this diagram. We claim that $X = \operatorname{colim}_{n \in \mathbb{N}} X_n$. Let $(g_n: X_n \to Y)_{n \in \mathbb{N}}$ be any other cocone. Then it being a cocone implies directly that $g_n(a_n) = g_m(a_m)$ for all $n, m \in \mathbb{N}$, and similar for the $b_n$. So we only need to show that $g_1(a_1) = g_1(b_1)$. This is where the choice of maps comes in, using that each $g_n$ is non-expansive we have for every $n \in \mathbb{N}$ that:
$$
d(g_1(a_1), g_1(b_1)) =
d(g_n(a_n), g_n(b_n)) \leq
d(a_n, b_n) = \frac{1}{n}
$$
So we must have that $d(g_1(a_1), g_1(b_1)) = 0$ and thus $g_1(a_1) = g_1(b_1)$.
Now let's finish the argument. Let $Z$ be any non-empty metric space. There is a (unique) map $h: Z \to X$ with constant value $*$. This map factors through the diagram, but not essentially uniquely. To see this, consider the constant map $h_1: Z \to X_1$ with value $a_1$, and the constant map $h'_1: Z \to X_1$ with value $b_1$. Both of these give factorisations of $h$. Since $Z$ is non-empty, these are different maps and there is no $f_{1n}: X_1 \to X_n$ such that $f_{1n} h_1 = f_{1n} h'_1$.
So the only place where non-expansiveness was used, was in the calculation of the colimit. If we allow any (continuous) function then there is, up to isomorphism, a unique two-point space. In particular, every map $f_{nm}: X_n \to X_m$ is an isomorphism. So the colimit of $(X_n)_{n \in \mathbb{N}}$ is just the two-point space again.
