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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?

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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.

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