FALSE: every continuous self-map of the circle is nullhomotopic

Suppose, for contradiction, that the identity is nullhomotopic. By [A1], there are $c\in \mathbb{R}\mathrm{/}\mathbb{Z}$ and a homotopy $H$ with $H(y,0)=y$ and $H(y,1)=c$. Reverse its time coordinate to obtain $K(y,t)=H(y,1-t)$, so $K(y,0)=c$ and $K(y,1)=y$.

Since $p$ is surjective, choose $a\in \mathbb{R}$ with $p(a)=c$. The constant map $y\mapsto a$ lifts $K(-,0)$, so [L1] and [L2] give a lift $\tilde{K}:(\mathbb{R}\mathrm{/}\mathbb{Z})\times I\to \mathbb{R}$. Define $s(y)=\tilde{K}(y,1)$. Then $s$ is continuous and $p(s(y))=K(y,1)=y$, so $p\circ s$ is the identity: $s$ is a section of $p$.

The path $s\circ {\omega }_1$ is a lift of ${\omega }_1$ because $p\circ s$ is the identity, and it is closed because ${\omega }_1(0)={\omega }_1(1)=[0]$. Put $m=s([0])\in \mathbb{Z}$ by [L5]. The path $t\mapsto m+t$ is another lift of ${\omega }_1$ starting at $m$, since [L4] and [L5] give $p(m+t)=[t]$; it is continuous by [L6]. Uniqueness in [L3] forces $s({\omega }_1(t))=m+t$, whose endpoint is $m+1\mathrm{\ne }m$, contradicting that $s\circ {\omega }_1$ is closed.

The contradiction discharges the assumption of step 1.1. Hence the identity is not nullhomotopic, and the universal claim is false.