Let $X$ be an arbitrary set equipped with the trivial topology $\{\emptyset, X\}$. Show that $\pi_1(X) = \{1\}$.

Question

I've just learned about homotopy and so many things are still confusing. I'd like to get some hints and explanation for this exercise. It is designed for me to get used to the basic concepts, so I'd like to keep the proof elementary.

Let $\alpha$ be a loop in $X$ with base point $p$, and $c_p$ be the constant loop at $p$. From the definitions, I assume that I will need to find a map $F: I \times I \to X$ such that $F(x,0) = \alpha(x)$ and $F(x,1) = c_p(x), \ \forall x \in I.$ Is this the right direction? How should I proceed after this?

Answer 1

If $X$ is equipped with the trivial topology, then any map from any topological space into $X$ is continuous. So, define $F$ so that both conditions that you stated hold and that will be it.

Answer 2

Stronger condition holds. If $x_0\in X$ then $\{x_0\}$ is a (strong) deformation retract of $X$ via

$$F:X\times I\to X$$ $$F(x,t)=\begin{cases} x &\text{if }t=0 \\ x_0 &\text{otherwise} \end{cases}$$

Thanks to the trivial topology $F$ is continuous.

In particular $X$ is contractible, hence $\pi_1(X)$ is trivial.