Fundamental groupoid of a contractible space

Question

I read that the fundamental groupoid of a contractible space is indiscrete. How can one show this? I found this as an exercise here.

Answer 1

If $f:(X,A)\to(Y,B)$ is a map of pairs of spaces, it induces a morphism $\pi f:\pi(X,A)\to\pi(Y,B)$ between their fundamental groupoids (with base points in $A$ and $B$, respectively). This actually gives a functor from the category of pairs of spaces to the category of groupoids.

Now if $F:(X\times I,A\times I)\to (Y,B)$ is a homotopy, this induces a functor $$ \pi(X,A)×\mathbf I\to\pi(X,A)\times\pi I\to \pi(X×I,A×I)\to\pi(Y,B) $$ Here $\mathbf I$ is tree groupoid with two objects, the fundamental groupoid of $I$ with base point $0$ and $1$. The middle map is a canonical isomorphism between the product of fundamental groupoids and the fundamental groupoid of the product.
A morphism $f:G\times\mathbf I\to H$ of groupoids is called a homotopy, with $\mathbf I$ playing for groupoids the same role as the unit interval does for spaces. Such a homotopy is basically a natural isomorphism between the restrictions of $f$ on $G\times\{0\}$ and $G×\{1\}$, with the arrows $f(1_x,0\to 1)$ as components.

Now let $X$ be a contractible space, and let $Y$ be the one-point space. We have maps $r:X\to Y$ and $i:Y\to X$ such that $ir$ is homotopic to $1_X$. Thus the identity on $\pi X$ is naturally isomorphic to a constant morphism $\pi X\to\{x_0\}$. From this it follows that each $\pi X(x,y)$ has precisely one element.