**Short version:** Yes, me.

**Slightly Longer Version:**

*Added in Edit*: What I write below is concerned with the space of *continuous piecewise-smooth* paths or loops. That is, continuous maps which are smooth except on a finite subset of the interval (or circle). Dropping the continuity requirement is only a cosmetic change: all the regularity statements shift down by one degree. Indeed, differentiation is an isomorphism from *continuous and piecewise-smooth* to just *piecewise-smooth* (providing you've chosen your end-point conditions compatibly). In particular, the analysis in the linked paper will readily adapt to this case.

There *is* a manifold of piecewise-smooth loops or paths, but it isn't pretty. It's just about as bad as you can get and still be called a manifold.

The first thing to decide is exactly what you mean by "piecewise-continuous smooth". The basic question is as to what you want a smooth map $[0,1] \to \mathbb{R}$ to be. Should it be smooth on $(0,1)$ and continuous on $[0,1]$ or should the derivatives exist at the end points?

If the first ("open"), then you're in for a nasty shock. The space of piecewise-smooth paths in $\mathbb{R}^n$ when considered as a locally convex topological vector space turns out to be a topological subspace of the space of continuous paths in $\mathbb{R}^n$. That's right: *topological* subspace. So all that information that you thought you had about derivatives is thrown away: you're using the $\|-\|_\infty$ norm. This means that any manifold structure that you are looking at is as a subspace of the Banach manifold of continuous paths (or loops). So since completeness is quite important, you should really work with continuous paths.

So then we look at the second ("closed"). In this case, the situation looks a bit better. What happens here is that adding additional breakpoints plays nicely: the space with breakpoints at $\{t_1,...,t_n\}$ is a closed subspace of the space with breakpoints at $\{t_1,...,t_n,t_{n+1}\}$. The difficulty here is that the family that you are taking the colimit over is uncountable (all finite subsets of the interval or circle) which means that all the "nice" theorems about inductive limits of LCTVS's don't (necessarily) hold.

Nonetheless, the colimit is a locally convex topological vector space and the space of piecewise-smooth loops or paths in an arbitrary manifold is a smooth manifold modelled on it. But that's the best you can say. I've not been able to show yet that it is paracompact, nor even that it is smoothly regular (like Tychanoff but with smooth functions).

However, there's a worse problem with this manifold. Let's take piecewise-smooth loops. Then the obvious property that you'd like is for the circle to act nicely on this space. Any given circle element acts by translation and this is a diffeomorphism. However, the assignment $\lambda \to \tau_\lambda$ is not continuous. The image of the circle is discrete. In fact, the image of the diffeomorphism group of the circle acting by reparametrisation is totally disconnected. So that's pretty bad as it means that all the standard reparametrisation homotopies don't work *as is*. They only work by homotoping the whole space to the space of smooth loops and working there - in which case you should probably work with smooth loops in the first place.

If you only allow breaks at *rational* points then life is a bit better in terms of the manifold's structure (it admits smooth partitions of unity, for example) but the circle action is still bad.

Fixing the circle action requires adding in lots more loops, right up to the space of differentiable loops with derivative of bounded variation. So that's almost as bad as the continuous loops that we had earlier.

Conclusion: either way, it's a Bad Space.

**Long Version:** Read The Smooth Structure of the Space of Piecewise-Smooth Loops.