Let $Y\subset S^{1}$ be a proper subspace of the unit circle. Prove that does not exists any homotopy $F$ between $\operatorname{id}_{S^{1}}$ and $i\circ f$, where $f: S^{1}\rightarrow Y$ is any continuous function and $i$ denotes the inclusion
Although I see that, I'm stuck with this exercise. By contraddiction, we would have a continuous function $$F:I\times S^{1}\rightarrow S^{1}$$ such that $$F(0,x) = x,$$ and $$F(1,x) = i(f(x))\in Y.$$ If $Y =\{ x_{0}\}$ we would have a retraction, hence it couldn't be since $S^{1}$ is not contractible. The problem come up when I consider the case $Y\neq \{ x_{0}\}$ and I can't see how to conclude. Can someone help me? Thanks before.
All you have to komw that $S^1$ is not contractible.
Since $Y$ is a proper subspace of $S^1$, there exists $Z \in S^1$ such that $Y \subset U_z = S^1 \setminus \{z\}$. The set $U_z$ is homeomorphic to an open interval, hence contractible.
Let $j : U_z \to S^1$ be the inclusion. Define $$f' : S^1 \xrightarrow{f} Y \hookrightarrow U_z .$$ Then $j \circ f' = i \circ f$.
Since $U_z$ is contractble, the map $j \circ f'$ is homotopic to a constant map, thus the same is true for $i \circ f$.
This excludes $i \circ f \simeq id$ because that would mean that $S^1$ is contractible.