# Continuous Left-inverse of Dirac Lipschitz-Free Space

Let $$X$$ be a separable pointed metric space and let $$AE(X)$$ denote the corresponding Lipschitz-Free (or Arens-Eells space) over $$X$$. The point-evaluation map $$\delta:X\mapsto AE(X)$$ is injective hence it has a left-inverse. When $$X$$ is Banach, the left-inverse may be taken to be continuous and linear.

In general, if $$X$$ is connected but not necessarily a topological vector space, does a continuous left-inverse of $$\delta$$ exist?

You're going to need that $$X$$ is contractible at least.
Let $$X$$ be the unit circle with the metric inherited from $$\mathbb{R}^2$$ and some point chosen as a base point. Assume that $$f$$ is a retraction of $$AE(X)$$ onto $$X$$. Since $$AE(X)$$ is a Banach space it is contractible. Let $$g:AE(X)\times[0,1]\rightarrow AE(X)$$ be a contraction of $$AE(X)$$ to any point. Then the function $$(x,t)\mapsto f(g(\delta(x),t))$$ is a contraction of $$X$$ to a point, but $$X$$ isn't a contractible space, so no such $$f$$ can exist.
I vaguely suspect that you need $$X$$ to be an absolute retract. Having a continuous left-inverse certainly implies that $$X$$ is an absolute retract relative to Lipschitz embeddings into Banach spaces.