# The status of $\mathbb{R}$ in homotopy theory.

The definition of a path as a continuous map $I \rightarrow X$ is a completely natural one. But this raises two questions in my mind. First, what properties of the interval give rise to useful topological invariants? Things like compact, metric, complete and ordered all seem fairly important but it's easy to ignore where they're being relied upon (precisely because the definition of path is so intuitive).

The obvious follow up question would be, can we replace $I$ with some other space and still develop an interesting theory? I guess you could go either way, either weakening or strengthening the definition of path.

• If you defined a path in $X$ as a continuous map from $\mathbb{R} \to X$, then how would you concatenate two such paths? Commented Jan 20, 2012 at 16:27
• This might be relevant: mathoverflow.net/questions/80777/what-is-a-continuous-path (I would like to make this a comment, but I don't have enough reputation.) Commented Jan 20, 2012 at 16:40
• (somewhat) related: mathoverflow.net/questions/28380/… Commented Jan 20, 2012 at 16:44
• @wckronholm So you're saying we need the domain to have nontrivial boundary, like for the higher homotopy groups? Those two links are great!
– dls
Commented Jan 20, 2012 at 17:00
• It think that's about as precise as I can be for a soft question. Like I said, the two links to MO questions are exactly what I was looking for. Here's another example I found this morning (though I understand this one even less): en.wikipedia.org/wiki/A%C2%B9_homotopy_theory
– dls
Commented Jan 20, 2012 at 20:25

Paths might be interesting but loops are far more interesting. That's because the multiplication on homotopy classes (induced by path composition) in case of paths is only associative but it generates a group structure in case of loops. So if we restrict to loops then there's a neat way to generalize the concept. First of all note that there's an equivalence between loops $I\to X$ and maps $S^1\to X$.

One of the fundamental properties of $S^1$ that is used is that it behaves well under wedge sums (as in gluing two spaces at given points).

Generally, let $X$ be a topological space, $x_0\in X$ a fixed point and

$$\psi: X\to X\vee X$$

a map such that right side is glued in $x_0$ (in both sides) and $\psi(x_0)=x_0$ (which is unique in $X\vee X$).

Now let $Y$ be another topological space with a fixed point $y_0\in Y$. Consider $f, g:X\to Y$ continous such that $f(x_0)=g(x_0)=y_0$, then we can define

$$f\cdot g : X\to Y$$ $$f\cdot g = (f\vee g)\circ\psi$$

And the new map again maps $x_0$ to $y_0$. You may think of $(\cdot)$ operator as a generalized path composition.

Now the essential assumption on $X$ is that if we now move to the set of homotopy equivalences of $\mathcal{C}((X, x_0), (Y, y_0))$ (i.e. the space of all continous maps $h:X\to Y$ such that $h(x_0) = y_0$), then

$$[f] \cdot [g] := [f\cdot g]$$

should define a group multiplication on the homotopy equivalences.

In the case of a sphere $S^n$ the example of such $\psi$ is (informaly) given by a contraction of the equator to a point. The resulting space can be easily identified with $S^n\vee S^n$. And for a given pointed space $(Y, y_0)$ the construction above yields a group naturally isomorphic to homotopy groups $\pi_n(Y, y_0)$.

There is a catch in all of that though. Notice how I said that the induced mutliplication should define group operation on homotopy equivalences. And what is a homotopy? It's a function $I\times X\to Y$. So at the fundamental level you always have $I$. The reason is it's just so damn good to model "fluent transformation from one function to another" with it.