Recently I have been reading the paper The categorical origins of Lebesgue integration by Tom Leinster (https://arxiv.org/pdf/2011.00412.pdf). In this paper, he said that：
For $n \geq 0$, let $E_{n}$ be the subspace of $L^{p}[0,1], (1\leq p<\infty)$ consisting of the equivalence classes of step functions constant on each of the (open) intervals $\left(\frac{i-1}{2^{n}}, \frac{i}{2^{n}}\right)$, $\left(1 \leq i \leq 2^{n}\right)$. Write $E=\bigcup_{n \geq 0} E_{n}$, which is the space of step functions whose points of discontinuity are dyadic rationals. Then $E$ is dense in the set of all step functions on $[0,1]$, which in turn is dense in $L^{p}[0,1]$; so $E$ is dense in $L^{p}[0,1]$. It follows that $L^{p}[0,1]$ is the colimit (direct limit) of the diagram $E_{0} \hookrightarrow E_{1} \hookrightarrow \cdots$ in the category $\mathbf{\text{Ban}}$ of Banach spaces with linear contractions as morphisms.
I know that the category $\mathbf{\text{Ban}}$ is cocomplete, i.e., the colimit of any diagram exists (because it has coproducts and coequalizers). However, I just have no idea why the density of $E$ in $L^p[0,1]$ implies just that $L^p[0,1]$ is the colimit of the increasing sequence $E_{0} \hookrightarrow E_{1} \hookrightarrow \cdots$ in the category $\mathbf{\text{Ban}}$ of Banach spaces with linear contractions? I thought about this for two days but didn't figure out anything. Can anyone give a convincing or inspiring explanation?
Any help is appreciated.
Best Answer
A colimit is an object $E$ together with morphisms $i_n:E_n\to E$ commuting with the inclusions $i_{n,m}:E_n\to E_m$ (i.e., $i_m\circ i_{n,m}=i_n$) such that, for every sequence of morphisms $f_n:E_n\to Y$ with $f_m\circ i_{n,m}=f_n$, there is a unique $f:E\to Y$ with $f_n=f\circ i_n$ for all $n$.
You have to show that $i_n:E_n\hookrightarrow L^p([0,1])$ satisfy this universal property. Given $f_n:E_n\to Y$ as above you can define a linear contraction $\varphi: \bigcup_{n\in\mathbb N} E_n \to Y$ by $\varphi(x)=f_n(x)$ for $x\in E_n$ (which is well-defined) and, because of the density of the union in $L^p([0,1])$, define $f:L^p([0,1])\to Y$ as the unique continuous extension of $\varphi$.