Can we develop homotopy theory choosing another space instead of $[0,1]$? An homotopy between two continous maps $F,G:X\to Y$ is a continous map
$$\mathscr{H}:X\times [0,1]\to Y,$$
such that $\mathscr{H}(-,0)=F$ and $\mathscr{H}(-,1)=G$. I was wondering whether it was interesting to use other spaces instead of $[0,1]$ in the definition.
More explicitely, given a topological space $J$ and two disjoint subspaces $J_0$ and $J_1$, we could define an $(J,J_0,J_1)$-homotopy between $F$ and $G$ as a continous map
$$\mathscr{O}:X\times J\to Y,$$
such that $\mathscr{O}(-,j_0)=F$ for all $j_0\in J_0$ and $\mathscr{O}(-,j_1)=G$ for all $j_1\in J_1$.
I thought that clearly ,for this definition to be interesting, it should induce an equivalence relation on the set of continous functions from $X$ to $Y$ just as classical homotopies (i.e. $([0,1],\{0\},\{1\})$-homotopies).
Reflexivity is clearly not a problem. Symmetric and transitive properties may be a problem.
Is this interesting? Or is it just a funny thought?
 A: Consider triples $(J,J_0,J_1)$, where $J$ is a topological space and $J_0,J_1\subseteq J$ are disjoint, non-empty, closed subspaces. Then, we define a $(J,J_0,J_1)$-homotopy between two continuous maps $f,g\colon X\rightarrow Y$ of topological spaces to be a map $H\colon X\times J\rightarrow Y$, such that $H(-,j_0)=f$ for all $j_0\in J_0$ and $H(-,j_1)=g$ for all $j_1\in J_1$. If $(J,J_0,J_1)=([0,1],\{0\},\{1\})$, this is just the usual notion of homotopy. Obviously, $(J,J_0,J_1)$-homotopy is a reflexive relation on the space of maps $X\rightarrow Y$. Note if $J_0,J_1$ were not disjoint, this would be the identity relation and if $J_0$ or $J_1$ were empty, this would be the trivial relation, whence we exclude those cases.
The condition that $J_0,J_1\subseteq J$ be closed is slightly restrictive, but easily motivated: Assume that $Y$ is Hausdorff. Let $j\in\overline{J_0}$ and $(j_{\alpha})_{\alpha}$ be a net in $J_0$, such that $j_{\alpha}\rightarrow j$. Then, for each $x\in X$, $H(x,j_{\alpha})=f(x)$ on one hand and $H(x,j_{\alpha})\rightarrow H(x,j)$ on the other hand. By the uniqueness of limits in Hausdorff spaces, $H(x,j)=f(x)$. Since $x$ was arbitrary, this means $H(-,j)=f$ for all $j\in\overline{J_0}$. Analogously, $H(-,j)=g$ for all $j\in\overline{J_1}$. This means that a $(J,J_0,J_1)$-homotopy is automatically a $(J,\overline{J_0},\overline{J_1})$-homotopy and the converse holds trivially, so we may WLOG assume that $J_0,J_1\subseteq J$ are closed (it is not forced that $\overline{J_0}$ and $\overline{J_1}$ are disjoint, but if they aren't, we again just obtain the identity relation). In case the codomain isn't Hausdorff, the assumption may lose generality, but who cares about generalities of non-Hausdorff spaces.
Now, replace $(J,J_0,J_1)$ by the triple $(J/\sim,J_0/\sim,J_1/\sim)$, where $\sim$ is the relation on $J$ that identifies each $J_0$ and $J_1$ to a point respectively. Note that $J_0/\sim,J_1/\sim\subseteq J/\sim$ are closed points. Every $(J/\sim,J_0/\sim,J_1/\sim)$-homotopy induces a $(J,J_0,J_1)$-homotopy by pullback. Conversely, $J\rightarrow J/\sim$ is a closed map (here, we crucially need that $J_0,J_1\subseteq J$ were assumed closed), so that $X\times J\rightarrow X\times J/\sim$ is a closed map, whence a quotient map, for any topological space $X$. This implies that any $(J,J_0,J_1)$-homotopy factors through  a $(J/\sim,J_0/\sim,J_1/\sim)$-homotopy. Thus, we may assume WLOG that $J_0=\{j_0\}$ and $J_1=\{j_1\}$ are points.
Thus, we are working in the category whose objects are bipointed spaces $(J,j_0,j_1)$, i.e. $J$ is a topological space and $j_0,j_1\in J$ are distinct, closed points. A bipointed map $\varphi\colon(J,j_0,j_1)\rightarrow(K,k_0,k_1)$ is a continuous map $\varphi\colon J\rightarrow K$, such that $\varphi(j_0)=k_0$ and $\varphi(k_0)=k_1$. If there is a bipointed map $(J,j_0,j_1)\rightarrow(K,k_0,k_1)$, then every $(K,k_0,k_1)$-homotopy induces a $(J,j_0,j_1)$-homotopy by pullback, when $(J,j_0,j_1)$-homotopy is a coarser relation than $(K,k_0,k_1)$-homotopy.
The category of bipointed spaces has a relevant endofunctor. If $(J,j_0,j_1)$ is a bipointed space, consider the quotient $J\vee J=J\sqcup J/\sim$, where $\sim$ is the equivalence relation generated by $(j_1,0)\sim(j_0,1)$ (in this notation, we interpret the disjoint union as $J\sqcup J=J\times\{0\}\cup J\times\{1\}$). This yields a bipointed space $(J\vee J,[j_0,0],[j_1,1])$. The construction is functorial.
Let's say a coalgebra (in this context) is a bipointed space $(J,j_0,j_1)$ together with a bipointed map $(J,j_0,j_1)\rightarrow(J\vee J,[j_0,0],[j_1,1])$. A morphism of coalgebras is a bipointed map between the underlying bipointed spaces compatible with those given bipointed maps. Note that $([0,1],0,1)$ is a coalgebra with the bipointed map $([0,1],0,1)\rightarrow([0,2],0,2)$ given by multiplication with $2$ (here, we identified $[0,1]\vee[0,1]\cong[0,2]$). Now, it is a theorem of Peter Freyd that this object is terminal in the category of coalgebras. A reference is Theorem 2.2 in Tom Leinster's A general theory of self-similarity. The statement is actually fairly intuitive, but it's explained there and I'm already writing too much.
Now, assume that $(J,j_0,j_1)$-homotopy is a transitive relation. Consider the $(J,j_0,j_1)$-homotopy $\{j_0\}\times J\rightarrow J\vee J,\,(j_0,j)\mapsto[j,0]$ and the $(J,j_0,j_1)$-homotopy $\{j_0\}\times J\rightarrow J\vee J,\,(j_0,j)\mapsto[j,1]$. The first proves that the maps $j_0\mapsto[(j_0,0)]$ and $j_0\mapsto[(j_1,0)]$ and the second proves that the maps $j_0\mapsto[(j_0,1)]$ and $j_0\mapsto[(j_1,1)]$ are $(J,j_0,j_1)$-homotopic, all as maps $\{j_0\}\rightarrow J\vee J$. Taking into account that $[(j_0,1)]=[j_1,0]$ (this is precisely how we glued) and that the relation of $(J,j_0,j_1)$-homotopy was assumed transitive, there exists a homotopy $H\colon\{j_0\}\times J\rightarrow J\vee J$, such that $H(j_0,j_0)=[j_0,0]$ and $H(j_0,j_1)=[j_1,1]$, i.e. $H(j_0,-)\colon J\rightarrow J\vee J$ is a bipointed map. In other words, if $(J,j_0,j_1)$-homotopy is a transitive relation, $(J,j_0,j_1)$ supports the structure of a coalgebra, in particular the theorem of Freyd implies the existence of a bipointed map $(J,j_0,j_1)\rightarrow([0,1],0,1)$.
Summa summarum, if we assume that $(J,J_0,J_1)$-homotopy is a transitive relation, then it is a coarser relation than regular homotopy. Note that if we further assume that $J$ is path-connected, then there is a bipointed map $([0,1],0,1)\rightarrow(J,J_0,J_1)$, so in that case $(J,J_0,J_1)$-homotopy is the same equivalence relation as regular homotopy (in particular, automatically symmetric). This is perhaps quite surprising.
