[Math] How to show the constant path is the identity element in fundamental group

algebraic-topologycontinuityfundamental-groupsgeneral-topologyhomotopy-theory

Let $X$ be a topological space and $q$ is a point in $X$. Denote the fundamental group of $X$ based at $q$ by $\pi _1(X,q)$. Then how should I verify that the constant path $c_q(s)\equiv q$ is the identity element? Here is my attempt:

Let $f$ be any path class in $\pi _1(X,q)$.
Then $$(f\cdot c)(s) = \begin{cases} f(2s) & {0\le s\le \frac 12 } \\ q & {\frac 12 \le s \le 1} \end{cases} $$

Define the map $H:I\times I\to X$ $$H(s,t)=\begin{cases} f(\frac {2s}{1+t}) & {0\le s\le \frac {1+t}2 } \\ q & {\frac {1+t}2 \le s \le 1} \end{cases}$$

Then the problem is how to show that this is continuous and hence it is actually a path homotopy.

The similar problem occurs when I want to show that a reverse path is the inverse element: I can write down a map that looks like a homotopy, but I got stuck at showing that it is continuous.

Best Answer

The domain is a union of two closed intervals that intersect at a point. All you need to know is that the two functions whose domains are those intervals are continuous and the two functions agree at that intersection point. In that case the function you get by piecing them together is continuous because the inverse image of a closed set is closed.

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