## Franklin's Notes

### Pseudocircle

The pseudocircle is the finite topological space on the four-point set $X={1,2,3,4}$ with the following open sets: We can visualize it as follows:

or we can visualize it using the Hasse diagram of its lattice of open sets:

Interestingly, the fundamental group of the pseudocircle is nontrivial, and in fact homotopy-equivalent to the unit circle $\mathbb S^1$. Consider the map $f:\mathbb S^1\to X$ sending the point with $\theta=0$ to $3$, the point with $\theta=\pi$ to $4$, all points with $\theta\in (0,\pi)$ to $1$, and all points with $\theta\in (\pi,2\pi)$ to $2$. We can visualize this mapping as follows:

We may also define a mapping $g:X\to \mathbb S^1$ sending To confirm that $f:\mathbb S^1\leftrightarrow X:g$ comprise a homotopy equivalence , we must show that $f\circ g\sim 1_X$ and $g\circ f\sim 1_{\mathbb S^1}$. The former map is just the identity map, making it trivially homotopic to the identity map, but the latter is not an identity map, so we must find a homotopy $H$ between $g\circ f$ and $1_{\mathbb S^1}$. This homotopy needs to be a continuous function $H:\mathbb S^1\times [0,1]\to \mathbb S^1$ such that $H(\theta,0)=g\circ f(\theta)$ and $H(\theta,1)=\theta$. I haven't yet managed to find such a function, but the fact that $f\circ g = 1_X$ is sufficient to at least show that the fundamental group of $X$ is nontrivial by the functoriality of $\pi_1$, the first fundamental group. In particular, since $f:\mathbb S^1\to X$ is split epic, we have that the group homomorphism $\pi_1 f:\pi_1(\mathbb S^1)\to \pi(X)$ is split epic as well. And since $\pi_1(\mathbb S^1)\cong\mathbb Z$, we have that $\pi(X)$ contains a subgroup isomorphic to $\mathbb Z$, which is sufficient to show nontriviality.

So if this space has a nontrivial fundamental group, what exactly is an example of a non-null-homotopic path? One example is the path $p:[0,1]\to X$ given by which, following the above diagram of how $\mathbb S^1$ maps to the unit circle, corresponds exactly to one clockwise rotation about the circle. That is, if $q:[0,1]\to\mathbb S^1$ is defined by $q(t)=e^{-2\pi i t}$, then we have that $p = f\circ q$.