# Could anyone suggest me a counter example about liftings?

Refer to the proof of the following assertion: Given a map $F\colon Y \times I \to S^1$ and a map $\tilde{F}\colon Y \times\{0\} \to \mathbb{R}$ lifting $F| Y\times\{0\}$, there is a unique map $\tilde{F}\colon Y \times I \to \mathbb{R}$ lifting $F$ and restricting to the given $\tilde{F}$ on $Y\times\{0\}$.

(a) Explain how the existence of a lift $F| Y\times\{0\}$ is used in the proof.

(b) Find an example of where the assertion fails if the existence of a lift of $F| Y\times\{0\}$ is not assumed.

What I am concerned about is the (b). Could anyone suggest me an example? I can't think of one myself.

• I cannot see the picture; could you perhaps type in whatever might be in it? – user122283 Sep 17 '14 at 16:17
• oh the picture doesn't come up? – Keith Sep 17 '14 at 16:32
• Given a map F : Y x I -> S1 and a map F' : Y x {o} -> R lifting F/(Y x {o}), there is a unique map F' : Y x I -> R lifting F and restricting to the given F on Y x {o}. (b) Find an example where the assertion fails if the existence of a lift of FjY f0g is not assumed. – Keith Sep 17 '14 at 16:35
• There must be a map $\Bbb R\to S^1$ in order to talk about lifts. I suppose it is the covering map $t\mapsto e^{2\pi it}$, but in order to make your question self-contained you should include that into your post. – Stefan Hamcke Sep 17 '14 at 18:38

The problem is uniqueness: there may be multiple lifts (and in fact there is always an infinite number of lifts if $Y \neq \emptyset$). For example if $Y$ is a point and $I \to S^1$ is a constant map, every point in the fiber over the image (something that looks like $\mathbb{Z}$ if the covering map is $t \mapsto e^{2\pi it}$) is a possible lift.