homotopy of paths with general interval

Question

I'm trying to understand a homotopy between two path. the definition of path in Topological space $X$ from point $x$ to $y$ is a continues map $f:[0,1]\rightarrow X$ such that $f(0)=x$ and $f(1)=y$.

A homotopy between to paths in Topological space $X$ from point $x$ to $y$; $f:[0,1]\rightarrow X$ and $g:[0,1]\rightarrow X$ is a continuous map $H:[0,1] \times [0,1]\rightarrow X$ such that $H(0,x)=f(x)$, $H(1,x)=g(x)$, $H(t,0)=x$ and $H(t,1)=y$.

How can I define the homotopy of paths with general interval like $[0,a]$ ? Assume we have two paths in Topological space $X$ from point $x$ to $y$; $f:[0,a]\rightarrow X$ and $g:[0,b]\rightarrow X$ what is the neutral way to define ?

Answer 1

In general for any topological spaces $X,Y$ and continuous maps $f,g:X\to Y$, a homotopy from $f$ to $g$ is a continuous map $H:X\times[0,1] \to Y$ such that $H(x,0)=f(x)$ and $H(x,1)=g(x)$ for all $x\in X$.

This generalizes your definition of a homotopy between paths. To do it for a general interval you can let $X=[a,b]$, but this is really the same definition as doing it for $X=[0,1]$ by just choosing a homeomorphism $h:[0,1]\to[a,b]$ and reparametrizing.

Note however that homotopies only make sense between maps who has the same domain and codomain. So a homotopy from a map $f:[0,a]\to X$ to a map $g:[0,b]\to X$ doesn't really make sense since they don't have the same domain. However, if you reparametrize such that the domains become the same, you can make sense of a homotopy between them. In other words, it might be useful to choose homeomorphisms $h_f:[0,1]\to [0,a]$, $h_g:[0,1] \to [0,b]$ and consider homotopies between $f\circ h_f$ and $g \circ h_g$.

Answer 2

I don't get the appeal of defining paths via general intervals. Anyway, one could potentially define a generalized notion of homotopy between a path $\gamma:[a,b] \rightarrow X$ and $\gamma':[a',b']\rightarrow X$ in the following way: Let $$\begin{align*} \square(a,b\mid a',b') &= \operatorname{conv} \{(0,a),(0,b),(1,a'),(1,b')\}\\ &=\{t_1(0,a) + t_2(0,b) + t_3(1,a') + t_4(1,b') \mid t_i\geq 0 \forall i, t_1+t_2+t_3+t_4=1\}\\ &\subseteq\mathbb R^2 \end{align*}$$ be the convex hull of the four points. You can depict it as something looking like $$\begin{array}{ccccc}(a',1) & - & (b',1)\\ &\diagdown&&\diagdown\\ &&(a,0) &-&(b,0) \end{array}$$ or $$\begin{array}{ccccccc} &&(a',1)&-&(b',1)\\ &\diagup&&&&\diagdown\\ (a,0)&-&---&-&---&-&(b,0) \end{array}$$ etc. Then a homotopy from $\gamma$ to $\gamma'$ can be defined as a map $$h:\square(a,b\mid a',b')\rightarrow X$$ such that $$h\vert_{\{0\}\times[a,b]} = \gamma\;\;\text{ and }\;\;h\vert_{\{1\}\times[a',b']}=\gamma'$$ If you want a homotopy relative endpoints $x$ and $y$, just add the requirement that $$h\vert_{((1-t)a+ta',t)} = x \;\;\text{and}\;\;h\vert_{((1-t)b+tb',t)}=y$$ for all $t\in [0,1]$.

I've never seen this construction being used somewhere though. Maybe it is, because it is unnecessary hustle and just using the unit interval makes things easier. For example it is not immediately clear, how to show that this "new" notion of homotopy defines an equivalence relation (of cause it is)...