What does it mean to lift a curve $\gamma$ in $\mathbb{C}\mathbb{P}^n$?

Question

Given a curve $\gamma$ in $\mathbb{C}\mathbb{P}^n$, what does it mean to lift it to $\mathbb{C}^{n+1}$? Is this lift a curve $\widetilde{\gamma}$ such that $\gamma = \iota \circ \widetilde{\gamma}$ and $\iota$ is isomorphism $\mathbb{C}\mathbb{P}^n \cong \mathbb{C}^{n+1} \setminus \{0\}$ quotient by action of the multiplicative group of nonzero complex numbers? It's also possible to lift discrete polygons in $\mathbb{C}\mathbb{P}^n$ in the same way, correct?

I came across this in a paper, but unfortunately I know very little about (complex) projective geometry. Any suggestion on where I can learn more about this (especially curves/polygons in projective spaces) would be great. Thanks!

Edit: Answering this question (after asking advisor): lifting a curve in this context simply means picking a representative in the space of homogenous coordinates $\mathbb{C}^{n+1}$. The accepted answer is not wrong, but continuity is completely irrelevant here.

Answer 1

The answer I previously accepted to this question is not correct. Lifting a curve in $\mathbb{C}\mathbb{P}^n$ to $\mathbb{C}^{n+1}$ in the context of projective geometry simply means picking a representative of homogeneous coordinates $[x_0 : \dots : x_n], x_i \in \mathbb{C}$ as stated in the second sentence question.

Answer 2

You probably meant a continuous curve $\gamma(t)\in \Bbb{P^n},t\in [0,1]$ in that case a lift always exists : locally a lift exists (just take one of the non-zero projective coordinate constant) so we get a continuous curve $\phi(t)\in \Bbb{C^{n+1}},t\in [0,1]$ such that $[\phi(t)]=\gamma(t)$,

if $\gamma(0)=\gamma(1)$ then $\phi(1) = a \phi(0),a\in \Bbb{C^*}$ and $\tilde{\gamma}(t)=\phi(t) e^{-t \log a}$ is a loop too.