# piecewise continuously differentiable path homotopy with fixed endpoints

Let $$U$$ be an open set in $$\mathbb{C}$$. Let $$f_0, f_1:[a,b] \rightarrow U$$ be (path) homotopic piecewise continuously differentiable functions with a homotopy $$H$$ and the additional property $$f_0(a) = f_1(a)$$, $$f_0(b) = f_1(b)$$. Is it possible to find a piecewise continuously differentiable homotopy $$\tilde{H}:[a,b] \times [0,1] \rightarrow U$$ between $$f_0$$ and $$f_1$$ such that $$(*) \quad \tilde{H}(a, t) = f_0(a) = f_1(a),\quad \tilde{H}(b, t) = f_0(b) = f_1(b)$$ for all $$t \in [0,1]$$?

We know that finding a homotopy without ensuring the property $$(*)$$ is possible, since we can use the compactness of $$H([a,b] \times [0,1])$$ to find an $$\epsilon > 0$$ such that the $$\epsilon$$-neighbour of $$H([a,b] \times [0,1])$$ is contained in $$U$$. And from there we can just approximate $$H$$ using another continuously differentiable function $$H'$$ to the point that $$\sup_{s,t}||H(s,t) - H'(s,t)|| < \epsilon.$$ Lastly, we glue the homotopy $$H'$$ with some linear homotopes from $$f_0(t)$$ to $$H(t, 0)$$ and $$f_1(t)$$ to $$H(t,1)$$. Hence we get a continuously differentiable homotopy but still likely without the property $$(*)$$.

How do we find $$\tilde{H}$$? Is it possible to modify $$H$$ to get $$\tilde{H}$$? Or do we need another approach?

• What assumptions are given regarding the domain and range of $H$ in your post? In the context of your first two sentences it seemed clear that $H : [a,b] \times [0,1] \to U$, but then the later paragraph seems not to assume that. However, if you simply meant $H : [a,b] \times [0,1] \to \mathbb C$ then I would say that $H$ plays little role in your problem, because all paths in $\mathbb C$ are homotopic in $\mathbb C$. So it's hard to figure out what you're really asking... Apr 29, 2023 at 15:15
• @LeeMosher I was talking about $U \subset \mathbb{C}$. Not all paths are homotopic in $\mathbb{C}$ if you also give conditions on the endpoints in order to call two paths homotopic Apr 29, 2023 at 16:42
• In that case you should hit the edit button to clear up two things in your post: to make the domain and range of $H$ explicit; and to emphasis the path homotopy concept which you did not mention in your post. Apr 29, 2023 at 17:57
• @LeeMosher the domain was always explicit, and in the post its said that $H$ is a homotopy with an additional restrictions on the endpoints $(*)$ which is the definition of path homotopy. I'm sorry I didn't mention the term path homotopy, I simply never used it, but will start to from now on. Apr 29, 2023 at 18:18
• Well, I'll try once more: is the homotopy $H$ given as $H : [a,b] \times [0,1] \to \mathbb C$? Or is it instead given as $H : [a,b] \times [0,1] \to U$? Apr 29, 2023 at 18:37

1. If you first arrange to satisfy condition (*) in the category of continuous homotopies in $$\Omega$$, you get a continuous map on $$D=[0,1]\times S^1$$ into $$\Omega$$ which by uniform continuity can be sampled on a triangular grid on $$D$$ and interpolated by a piece-wise linear map that still takes values in $$\Omega$$. Note that this map will satisfy the condition of being piece-wise differentiable. If you want the map to be even smoother than that and explictly constructed, you can look at splines (e.g. piecewise-polynomial maps.)