The Fundamental group of the circle from "Introduction to knot theory", Ralph H. Fox (1')

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The book says in the beginning of discussing this title:

"Let the field of real numbers be denoted by R and the subring of integers by $J$. We denote the additive subgroup consisting of all integers which are a multiple of 3 by $3J$. The circle, whose fundamental group we propose to calculate, may be regarded as the factor group $R/3J$ with the identification topology, i.e., the largest topology such that the canonical homomorphism $\phi: R \rightarrow R/3J $ is a continuous mapping.A good way to picture the situation is to regard $R/3J$ as a circle of circumference 3 mounted like a wheel on the real line $R$ so that it may roll freely back and forth without skidding. the possible points of tangency determine the $many-one$ correspondence $\phi$"

And then the book started proving the following:

The image under $\phi$ of any open subset of $R$ is an open subset of $R/3$ (5.2)

And the proof is given below:

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My Question:

I do not understand why $\phi^{-1} \phi (X)$ is given by this form, could anyone explain this for me please?

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I do not understand what you mean exactly by $\phi^{-1} \phi (X)$ is given in this form but if you would like some intuition on its use in the proof, I can provide that.

To show the proposition "the image under of any open subset of is an open subset of /3 (5.2)" is true it is sufficient to show that the inverse function is continuous. Our goal is now to show this using what we know to be true about the function phi, namely:

$\phi^{-1}\phi(X)=\bigcup_{n\in J}(3n+X)$

We know this to be true since the a little rewriting of the question gives us:

$\phi(X)=\phi\left(\bigcup_{n\in J}(3n+X)\right)$

and all this does is notes the fact that anything of the form $3n, n\in J$ is in the kernel of $\phi$. This makes sense since if we use the wheel analogy given by Fox anything that is a multiple of 3 is in the same equivalence class as zero in the image of the map due to the fact after the wheel rolls a distance of 3 units, it has returned back to zero.

Now once we have established this notion, it is clear to see that $\phi$ takes an open set $X$ and translates it to a collection of open sets $3n+X, \ n\in J$. We know the union of open sets is open and the conclusion follows as stated by Fox. The crux is now that since for a subset $B$, $B$ is open iff $\phi^{-1}(B)$ is open we have now shown $\phi^{-1}(B)$ to be open and it follows that $B$, aka $\phi(X)$ is also open.