I'm having trouble answering what seems to be a simple question that I will describe below.
Let $A$ be a topological space, where $x$ and $y$ are in A, and $P$ be a path from $x$ to $y$. The constant path from $[0, 1] \rightarrow A$ is defined as $c(s) = x$. The question asks to show that the concatenation of $c \star P$ is homotopic to $P$. This intuitively seems trivial, as a constant path of A means that for all $x \in A$ we get $c(s) = x$ for all $s \in [0,1]$. I have the following written down, but it doesn't feel correct. Any suggestions?
$H = \begin{cases} c(2s), & 0 \leq s \le \frac{1}{2} \\
P(2s-1), & \frac{1}{2} \leq s \le 1
\end{cases}$
Best Answer
You have written out the concatenation path, but not the homotopy yet, which is a function of two parameters.
$$H(s,t)=\begin{cases} P((1-t)s) & 0 \le s \le \frac{1}{2}, t \in [0,1]\\ P(st-t+s) & \frac{1}{2}\le s \le 1, t \in [0,1]\\ \end{cases}$$
Then $P(\frac12, t) = P(\frac12(1-t))$ from the first condition, and $P(\frac12, t)=P(\frac12 t - t + \frac12)$, both of which are $P(\frac12- \frac12 t)$, so it's always well-defined, and so clearly continuous. For all $t$ we also have $H(0,t) = P(0)=x$ and $H(1,t)=P(1)=y$ so we have a path from $x$ to $y$ for all $t$, so $H$ fixes the endpoints.
For $t=1$ we get the definition of $c \star P$, and for $t=0$ we have just $P$, so $P \simeq c \star P$.
In this note (page 4) I give some explanation (from reparametrisation) for this $H$, and prove all the homotopy group axioms, in quite a lot of detail.