understanding inverse image in topology the exercise I  was reading was finding one-point compactification of $(0, 1)$. the solution convinced me that was $S^1$ 
The problem is ths: It seems intuitive that $f^{-1}(V)$ is the set $Y-[\epsilon,1-\epsilon]$ where $Y$ is one-point compactification set. but how rigorously? can you please explain all detail ?
 A: We have
$$f^{-1}(V) = Y \setminus (Y \setminus f^{-1}(V)) = Y \setminus (f^{-1}(S^1) \setminus f^{-1}(V)) = Y \setminus (f^{-1}(S^1 \setminus V)) .$$
Therefore it suffices to show that $f^{-1}(S^1 \setminus V) = [\epsilon,1-\epsilon]$. Clearly $(1,0) \notin S^1 \setminus V$, thus $f^{-1}(S^1 \setminus V)$ is a subset of $(0,1)$. More precisely
$$f^{-1}(S^1 \setminus V) =  \{ t \in (0,1) \mid (\cos(2\pi t),\sin(2\pi t)) \in S^1 \setminus V \} = \{ t \in (0,1) \mid (\cos(2\pi t),\sin(2\pi t)) \notin V \} .$$

*

*For $t \in (0,\epsilon)$ we have $(\cos(2\pi t),\sin(2\pi t)) \in V$.


*For $t \in (1-\epsilon,1)$ we have $t-1 \in (-\epsilon,0)$, thus $2\pi t = 2\pi (t-1) + 2\pi$ and therefore $(\cos(2\pi t),\sin(2\pi t)) = (\cos(2\pi (t-1)),\sin(2\pi (t-1))) \in V$.


*For $t \in [\epsilon,1-\epsilon]$ we have $(\cos(2\pi t),\sin(2\pi t)) \notin V$. Otherwise
$$(\cos(2\pi t),\sin(2\pi t))  = (\cos(2\pi s),\sin(2\pi s)) \tag{*}$$ for some $s \in (-\epsilon,\epsilon)$. But (*) implies that $2\pi t$ and $2\pi s$ differ by an integral multiple of $2\pi$ which is impossible for  $t \in [\epsilon,1-\epsilon]$ and $s \in (-\epsilon,\epsilon)$.
