Making a path in a manifold differentiable I have a closed path in a smooth compact Manifold $C \subset M$, that consists of straight pieces, so $C = \alpha \circ \beta \circ ... \circ \delta$, where $\circ$ is the concatenation, and $\alpha = tx + (1-t)y$ , $\beta = ty+(1-t)z ,...$  for some $x,y,z,... \in M$ and $t \in [0,1]$. How would I change this path, to make it differentiable, assuming i dont have any self-intersections? I can choose a small neighbourhood of each of these $x,y,z,...$ where i can do whatever i want, to make it differentiable, and not change the homotopy. but how exactly do i change it, to make it differentiable?
 A: In my answer I'm going to use $*$ as the concatenation operator, since I must use $\circ$ in its usual role as the composition operator.
You can reparameterize each of the paths $\alpha,\beta,\ldots,\delta : [0,1] \to C$ by precomposing each with a homeomorphism $f : [0,1] \to [0,1]$ whose derivative $f'(t)$ is continuous and such that $f'(0)=f'(1)=0$. The resulting composed paths $\alpha \circ f,$ $\beta \circ f,$ ..., $\delta \circ f$ are homotopic rel endpoints to $\alpha,\beta,\ldots,\delta$ respectively, because $f$ is homotopic to the identity rel endpoints. The resulting concatenation $(\alpha\circ f) * (\beta\circ f) * \cdots * (\delta \circ f)$ is continuously differentiable, and it is homotopic rel endpoints to the original path $\alpha * \beta * \cdots * \delta$.
Straightness of the pieces $\alpha,\beta,\ldots,\delta$ is not necessary for this argument, and might be impossible to achieve for a general smooth compact manifold $C \subset M$. Fortunately, this exact same argument works as long as each of $\alpha,\beta,\ldots,\delta$ is continuously differentiable.
Also, the assumption of no self-intersections is not necessary for this argument, it works regardless of whether $\alpha \circ \beta \circ \cdots \circ \delta$ has self-intersections.
