loops $\alpha ,\beta$ based at $x_0$ are homotopic then is there another loop such that conjugating $\alpha$ with it is path-homotopic to $\beta$?

Question

Let $X$ be a topological space and $\alpha , \beta : [0,1] \to X$ be two loops based at $x_0 \in X$ . If $\alpha$ and $\beta$ are homotopic as maps (not path homotopic, only homotopic ) ; then does there exist a loop $\gamma : [0,1] \to X$ based at $x_0$ such that $\gamma *\alpha * \gamma ^{-1}$ is path-homotopic with $\beta$ ?

Answer 1

As you have stated the question, the answer is no, because $\alpha,\beta$ are always homotopic: $$H(s,t) = \begin{cases} \alpha(s(1-2t)) &\quad\text{if $0 \le t \le 1/2$} \\ \beta(s(2t-1)) &\quad\text{if $1/2 \le t \le 1$} \end{cases} $$

However, what you ask for is true if one states the homotopy requirement more carefully, namely that $\beta$ and $\alpha$ are homotopic through closed paths, although the base point is allowed to move during the homotopy, it is not required to stay fixed at $x_0$. This means that there is a homotopy of the form $h : [0,1] \times [0,1] \to X$ such that $h(s,0)=\beta(s)$, $h(s,1)=\alpha(s)$, and for all $t$ we have $h(0,t)=h(1,t)$. Then yes, what you ask for is true, and its pretty obvious using the path $\gamma(t)=h(0,t)=h(1,t)$.

Added in reponse to a comment: I'll outline how to construct a homotopy $H : [0,1] \to [0,1] \to X$ from $\beta$ to $\gamma\alpha\gamma^{-1}$, which is based on the given homotopy $h : [0,1] \times [0,1] \to X$. The formula for $H$ is a composition of the form $$H : [0,1] \times [0,1] \xrightarrow{g} [0,1] \times [0,1] \xrightarrow{h} X $$ The continuous map $g$ will satisfy a list of properties, and then one writes down a piecewise affine formula for $g$ which satisfies those properties (using skills learned in topology textbooks such as Munkres "Topology" or Hatcher's "Algebraic Topology"). Here are the properties that $g$ needs to have:

1. $g$ restricts to the identity on $[0,1] \times \{0\}$
2. $g$ takes $\{0\} \times [0,1]$ to $\{(0,0)\}$
3. $g$ takes $\{1\} \times [0,1]$ to $\{(1,0)\}$
4. $g$ takes $[0,1] \times \{1\}$ one-to-one onto $\bigl(\{0\} \times [0,1]\bigr) \cup \bigl([0,1] \times \{1\} \bigr) \cup \bigl( \{1\} \times [0,1] \bigr)$