Show $\prod_1(X,p)$ is non trivial; that is it contains an element besides the identity $[e_p]$

Question

Can someone please give me feedback/help with my proof for the problem below? Thank you.

Let $p,q\in X$, and let $f$ and $g$ be paths in $X$ from $p$ to $q$ for which $[f]\ne[g].$ Show $\prod_1(X,p)$ is non trivial; that is it contains an element besides the identity $[e_p]$.

$\textit{Proof.}$ Let $p,q\in X$, and let $f$ and $g$ be paths in $X$ from $p$ to $q$ for which $[f]\ne[g]$, i.e., $f$ and $g$ are not path homotopic. Our claim is $\prod_1(X,p)$ is non-trivial.

Let $[f*\bar{g}]$ be a path from $p\to p$, i.e., $f*\bar{g}\in \prod_1(X,p)$ on the contrary, if $\prod_1(X,p)$ is trivial then $[f*\bar{g}]=[e_p]$, i.e., $f*\bar{g}$ and $p$ are homotopic; there exists $f\colon I\times I \to X$ such that it is continuous and $f(s,0)=p$ for all $s\in I$, $f(s,1) = f*\bar{g}(s)$ for all $s\in I$, i.e., $f(s,1)=f(2s), s\in [0,v_2] \text{ and } \bar{g}(2s -1), s\in[y_2, 1].$

Then $G_1\colon I\times I \to X$ is defined by and $G_1(s,t) = f(\frac{s}{2},t)$ is continuous since $\frac{s}{2}$ and $t$ are continuous and $f$ is continuous. Also $G_1(s,0)=f(\frac{s}{2},0)=p$ for all $s\in I$ and $G_1(s,1)=f(\frac{s}{2},1)=f(s)$ for all $s\in I$. So $G_1$ is path homotopy between $f$ and $p$ say $G_2$. Similarly, there is a path homotopy between $\bar{g}$ and $p$ and then $g$ and $p$. Thus, $f$ and $g$ are both homotopic to $p$ implies $[f]=[g]$ which a contradiction to $[f]\ne[g]$ then $\prod_1(X,p)$ is non-trivial group.

Now, we define $G_2\colon I\times I \to X$ as $G_2(s,t)=f(\frac{1}{2}(s+1),t)$ is continuous since $f, \frac{s+1}{2},$ and $t$ are continuous. Moreover, $G_2(s,0) = f(\frac{1}{2}(s+1),0) = p$ for all $s\in I$ and $G_2(s,1) = f(\frac{(s+1)}{2},1)=\bar{g}(s)$ for all $s\in I$ is homotopy between $\bar{g}$ and $p$. Then take $G_3$ be homotopy between $g$ and $p$. So, $G_3\colon I\times I \to X$ defined by $G_3(s,t) = G_2(1-s,t).$

Answer 1

Because classes of paths under homotopy form a groupoid, essentially. If you know that you don't need the extra homeotopies:

So $[f] \neq [g]$ implies $[f \cdot \bar{g}] \neq [e_p]$, where $[e_p]$ is the class of the constant map with value $p$, for otherwise from $[f \cdot \bar{g}] = [e_p]$ we'd conclude $[f \cdot \bar{g}] \cdot [g] = [e_p] \cdot [g]$ and hence (transivity and right/left inverses etc) $[f]= [f \cdot (\bar{g} \cdot g)]=[g]$.

So it's a direct consequence of the groupoid properties.

Answer 2

Your idea is correct. To put it simply, one can concatenate things into the loop $h\colon [0,1] \to X$ given by $$h(t) = \begin{cases} f(2t)\mbox{ if }0 \leq t \leq 1/2 \\ g(2-2t) \mbox{ if }1/2 \leq t \leq 1 \end{cases}$$This $h$ is continuous because both $f$ and $g$ are and $f(1) = g(1) = q$. Also $h(0) = f(0) = g(0) = h(1)$ so $h$ is a loop. And $h$ is null-homotopic because $[f] \neq [g]$. Indeed, we have that $h = f \ast g^{\leftarrow}$ where $g^{\leftarrow}(t) = g(1-t)$ is $g$ traveled backwards, so $[h] = [f \ast g^{\leftarrow}]$. Then $[h] = [c_p]$ is equivalent to $[f] = [g]$ and you seem to have defined the relevant homotopies in detail.