Hatcher 1.3.27: Clarification on left and right actions This question has been asked here, and I am making a new question seeking clarification after a comment reply from Eric Wofsey.
I understand how both actions of $\pi_1(X,x_0)$ are defined on the fibre $p^{-1}(x_0)$, however I don't understand how the 'first action' (the action defined by lifting loops in $X$ at $x_0$ to points in the fibre) becomes a right action. As far as I can tell, it is a left action.
If I could get something more concrete for why this is true that would be an ideal answer.
 A: This is explained in Hatcher's book on p. 69. The elements of $\pi_1(X,x_0)$ are path homotopy classes of closed paths $\gamma : [0,1] \to X$ based at $x_0$ (i.e. $\gamma(0) = \gamma(1) = x_0$). Given $y \in  p^{-1}(x_0)$ and a path $\gamma$ based at $x_0$, we find a unique lift $\tilde \gamma$ of $\gamma$ such that $\tilde \gamma (0) = y$. Define $a(y, \gamma)  = \tilde \gamma (1)$. It is easy to verify that $\tilde \gamma (0)$ only depends on $[\gamma] \in \pi_1(X,x_0)$. That is, $a(y, [\gamma]) =  a(y, \gamma)$ is well-defined. The product of $[\gamma], [\delta] \in \pi_1(X,x_0)$ is given by $[\gamma] \cdot [\delta] = [\gamma  \cdot \delta]$, where the path $\gamma  \cdot \delta$ is defined by $(\gamma  \cdot \delta)(t) = \gamma (2t)$ for $t \le 1/2$ and $(\gamma  \cdot \delta)(t) = \delta (2t-1)$ for $t \ge 1/2$ (that is, we first travel with $\gamma$ and then with $\delta$). This shows that
$$a(y, [\gamma] \cdot [\delta]) = a(a(y, [\gamma]), [\delta]) .$$
In fact, we first lift $\gamma$ to $\tilde \gamma$ with $\tilde \gamma (0) = y$ to get $a(y, [\gamma]) = \tilde \gamma (1)$. Next we lift $\delta$ to $\tilde \delta$ with $\tilde \delta (0) = a(y, [\gamma]) = \tilde \gamma (1)$ to get $a(a(y, [\gamma]),[\delta]) = \tilde \delta (1)$. But the composed path $\tilde \gamma \cdot \tilde \delta$ is a lift of $\gamma \cdot \delta$ such that $(\tilde \gamma \cdot \tilde \delta)(0) = \tilde \gamma(0) = y$. Hence $a(y, [\gamma] \cdot [\delta]) = (\tilde \gamma \cdot \tilde \delta)(1) = \tilde \delta(1) = a(a(y, [\gamma]),[\delta])$. This shows that $a$ is a right action. See for example right group action. Writing suggestively $y  \cdot [\gamma] =  a(y, [\gamma])$ we get
$$y  \cdot ([\gamma] \cdot [\delta]) = (y  \cdot  [\gamma]) \cdot [\delta] .$$
This explains the name right action: The above "associative property" works via multiplication on the right side of $y$.
But isn't writing $y  \cdot [\gamma]$ quite arbitrary, coudn't we write $[\gamma] \cdot y$? Yes, we can, but then we get
$$([\gamma] \cdot [\delta]) \cdot y = [\delta] \cdot ([\gamma] \cdot y) $$
which doesn't look like a usual associative property.
This is the reason why Hatcher defines a left action by
$$[\gamma] * y = y \cdot [\gamma]^{-1} .$$
In fact
$$([\gamma] \cdot [\delta]) * y = y \cdot ([\gamma] \cdot [\delta])^{-1} = y \cdot ([\delta]^{-1} \cdot [\gamma]^{-1}) = (y \cdot ([\delta]^{-1}) \cdot [\gamma]^{-1} = ([\delta] * y ) \cdot [\gamma]^{-1} \\= [\gamma] * ([\delta] * y) .$$
