I saw this question, but the top answer has the same picture that is confusing me currently, so I was wondering if somebody could explain it to me. I'm reading Concepts of Modern Mathematics by Ian Stewart, and in it he is trying to define an inverse for paths so we can construct a group. He does this as follows:
If $p^{-1}$ is the path $p$ in reverse, then although $p*p^{−1}$ is not equal to the trivial loop, it is homotopic. As in Figure 139, we can shrink $p*p^{−1}$ gradually back towards the base point, at the same time going round it faster and faster.
Here's what I don't understand. Suppose $q = p*p^{-1}$:
- Are there only inverses for paths that start and end at the same point? What about a path that starts at point $a$ and ends at point $b$ such that $a \neq b$, this won't be homotopic to the trivial loop
- As $q$ is "shrinking", it is not covering the full distance, and instead of traveling from the initial point to some point that is not the initial point before returning. Doesn't this mean we're not dealing with another path, e.g. $r*r^{-1}$, in the second drawing, and another path in the third, etc. as it shrinks?
EDIT: Upon rereading the text, I realized that the author was trying to get to the fundamental group, which is why he works with loops. He explained homotopy by working with general paths, not just loops, and this is what got me confused since I didn't realize the general paths were only mentioned for this brief explanation, and most of the rest of the discussion involved loops. This document has a good explanation of why loops are used, saying that:
By restricting attention to loops with a fixed basepoint $x_0 \in X$, we guarantee that the product $fg$ of any two such loops is defined.
Best Answer
You're correct that you can talk about inverses without assuming that your paths are loops. Let me introduce some notation to clarify this. For $a, b \in X$ I will denote $p: a \longrightarrow b$ to mean that $p$ is a path starting at $a$ and ending at $b$. Given a path $p: a \longrightarrow b$, I will also denote its homotopy class as $[p]$. More specifically, to say $[p] = [q]$ I mean that there is a homotopy $H: I^2 \longrightarrow X$ such that $H(0, s) = p(s)$, $H(1, s) = q(s)$ for all $s$ and such that $H$ fixes the endpoints, so $H(t, 0) = a$ and $H(t, 1) = b$ for all $t$. This fixed endpoint condition means that for all $t$, the path $s \mapsto H(t, s)$ is a path $a \longrightarrow b$. Thus, it is well defined to say that $[p] : a \longrightarrow b$, as the fixed endpoint condition means that if $[p] = [q]$ then $q$ has the same endpoints $a \longrightarrow b$.
With that jargon aside, the point is that if we have paths $p: a \longrightarrow b$, $q: b \longrightarrow c$ then we can concatenate them to get $p*q: a \longrightarrow c$. We can also define the reverse path $p^{-1}: b \longrightarrow a$. Notably, concatenation is well defined on homotopy classes! That is, we can define the product $[p] * [q]$ as $[p * q]$ and have it be well defined. For $a \in X$, we will also denote $c_a: a \longrightarrow a$ as the constant path at $a$.
Now, with all of these definitions set up, we can prove that $[p] * [p^{-1}] = [c_a]$ for $p: a \longrightarrow b$. And we need not insist $a = b$ at all for this statement to make sense. The construction here is exactly as you mentioned for loops. As $[p] * [p^{-1}] = [p * p^{-1}]$ we are trying to show that $p * p^{-1}$ and $c_a$ are homotopic with fixed endpoints. Here, of course, $p * p^{-1} : a \longrightarrow b \longrightarrow a$ and $c_a: a \longrightarrow a$ so $a \longrightarrow a$ are our endpoints. For $t \in [0, 1]$, we will denote $p_t = p|_{[0, t]}$. Then $p_t: a \longrightarrow p(t)$. We also have its reverse $p_t^{-1} : p(t) \longrightarrow a$. Then we are claiming that $H(t, s) = (p_t * p_t^{-1})(s)$ is a homotopy between $p * p^{-1}$ and $c_a$ which fixes the endpoints $a \longrightarrow a$. Certainly, $H(0, s) = p_0 * p_0^{-1}(s) = c_a(s)$ and $H(1, s) = p_1 * p_1^{-1}(s) = p * p^{-1}(s)$. Furthermore, for any $t$, the path $s \mapsto H(t, s)$ is the path $p_t * p_t^{-1}: a \longrightarrow a$. Thus, $H$ does indeed fix endpoints so $H$ witnesses $[p * p^{-1}] = [c_a]$. Again, note that we never needed the fact that $a = b$. If you're interested more in this idea of paths as "maps" $a \longrightarrow b$, you could look into the fundamental groupoid. It is essentially the object I defined here consisting of things like $[p]: a \longrightarrow b$. Restricting this "fundamental groupoid" to only consider loops $[p]: a \longrightarrow a$ yields the fundamental group.
To answer your second question, certainly it is the case that the "shrunken" $p * p^{-1}$ are all completely different maps. Here, I wrote them as $p_t * p_t^{-1}$. However, they all have the same endpoints $a \longrightarrow a$ and essentially, the map $t \mapsto p_t * p_t^{-1}$ is a continuous family of loops at $a$, i.e. a homotopy.