a unitary does not in general lift to a unitary

Question

Consider the restriction map $\pi:C(\mathbb{D})\rightarrow C(\mathbb{T})$, where $\Bbb D$ is the closed unit disk and $\Bbb T$ is the unit circle. Suppose $v\in C(\Bbb T)$ is a unitary such that $v(z)=z$; why doesn't there exist a unitary $u\in C(\Bbb D)$ such that $\pi(u)=v$?

For anyone looking into this, $u$ as in the question cannot even be invertible. — Martin Argerami, Mar 18 '23 at 04:43

score 1 · Accepted Answer · answered Jul 27 '19 at 13:59

1

This is just a fact from topology: the identity map $v:\mathbb{T}\to\mathbb{T}$ cannot be extended continuously to a map $u:\mathbb{D}\to\mathbb{T}$. You can prove this using the fundamental group, for instance: $\pi_1(\mathbb{T})$ is nontrivial so $v$ induces a nontrivial map on fundamental groups and in particular is not nullhomotopic, but if it extended to $\mathbb{D}$ it would be nullhomotopic since $\mathbb{D}$ is contractible.

answered Jul 27 '19 at 13:59

Eric Wofsey

342,377

But $u$ is a map from $\Bbb D$ to $\Bbb C$. – math112358 Jul 28 '19 at 03:32
If it's unitary it must map to $\mathbb{T}$. – Eric Wofsey Jul 28 '19 at 03:35
oh,I see.Thanks! – math112358 Jul 28 '19 at 03:36

a unitary does not in general lift to a unitary

1 Answers1