The James construction $JA$ on a pointed type $A$. Is the higher inductive type given by

$\epsilon_{JA} : JA$

$\alpha_{JA} : A \to JA \to JA$

$\delta_{JA} : \prod_{x : JA} x = \alpha_A (*_A, x)$

Properties

We have an equivalence of types $JA \simeq \Omega \Sigma A$ if $A$ is 0-connected.

We can see that $JA$ is simply the free monoid on $A$. The higher inductive type is recursive which can make it difficult to study. This however can be remedied by defining a sequence of types $(J_n A)_{n: \mathbb{N}}$ together with maps $(i_n : J_n A \to J_{n+1} A)_{n:\mathbb{N}}$ such that the type $J_\infty A$ defined as the sequential colimit of $(J_n A)_{n:\mathbb{N}}$ is equivalent to $JA$.

This is useful as we can study the lower homotopy groups of $\Omega \Sigma X$ by studying the lower homotopy groups of the sequential colimit. Which is what allows Brunerie to show $\pi_4(S^3) = \mathbb{Z} / n \mathbb{Z}$.