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Consider the map $I:\mathbb{R}\to \mathbb{Z}, x \mapsto I(x) $ where $ I(x)$ is the integer part of $x$. On $\mathbb{R}$ we define an equivalence relation $xRy \Leftrightarrow I(x)=I(y)$. Then $I$ defines a bijection $\tilde{I}:\mathbb{R}/R \to \mathbb{Z}$. What is the topology on $\mathbb{Z}$ that makes $\tilde{I}$ an homeomorphism?

The map $I$ is not continuous since $\{1,2,3\}$ are open in $\mathbb{Z}$ endowed with subspace topology, but $I^{-1}\{1,2,3\}=[1,4)$ which is not open. Therefore, the standard topology on $\mathbb{Z}$ won't make $\tilde{I}$ an homeomorphism.

My intuition tells that it's the indiscrete topology. Is this correct?

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    The indiscrete topology will make the map continuous, but will it be a homeomorphism? You need to look at the open sets of $\Bbb R/R$ and see what they look like. – 2017-02-15
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    @Arthur open sets in $\mathbb{R}/R$ are empty set, $\mathbb{R}/R$ and $\pi (-\infty,n]$ where $n\in \mathbb{Z}$. Then how to make $\tilde{I}$ continuous? – 2017-02-15

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Hint Let $\pi : \Bbb R / R$ denote the quotient map, and suppose $U \subseteq \Bbb R / R$ is open. By definition, this is the case iff $\pi^{-1}(U)$ is open.

The key idea here is that each fiber of $\pi$, say, $[n, n + 1)$, respectively contains the point $n$ which is close to points in other fibers, respectively, $[n - 1, n)$, and so this is reflected in the topology $\pi$ determines on $\Bbb R / R$, which we want to identify with $\Bbb Z$ via a homeomorphism.

More precisely: Suppose $U$ is nonempty, say it contains $\pi(n)$ for some $n \in \Bbb R$; by definition of $R$ we may replace $n$ with its floor and hence assume $n$ is an integer. Now (assuming $\Bbb R$ is endowed with the standard topology) $\pi^{-1}(U)$ contains an interval $(n - \epsilon, n + \epsilon)$ for some $\epsilon > 0$ (and we may as well take $\epsilon < 1$). Thus, our open set $U$ must also contain $\pi(n - \epsilon) = \pi(n - 1)$. What does induction now tell us?

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    open sets in $\mathbb{R}/R$ are empty set, $\mathbb{R}/R$ and $\pi(-\infty,n]$ for $n\in \mathbb{Z}$. – 2017-02-15
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    Then I see with the standard topology on $\mathbb{Z}$, $\tilde{I}^{-1}$ is continuous. But $\tilde{I}$ is not. – 2017-02-15
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    That sounds right to me! Notice that it remains to show that $\pi((-\infty, n])$ is actually an open set. (One could do this by writing $\pi((-\infty, n])$ as $\pi(V)$ for a suitable $\pi$-saturated open set $V$.) – 2017-02-15
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    To show $\tilde{I}$ is an homeomorphism, we need to show that $\tilde{I}$ and its inverse are continuous. Then to show $\tilde{I}$ is continuous, we need to show for every open $U\subset \mathbb{Z}$, $\tilde{I}^{-1}(U)$ is open. – 2017-02-15
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    I don't know how to choose a right topology for $\mathbb{Z}$ that makes above happen. – 2017-02-15
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    Your first comment above identifies what (I believe) is the correct topology on $\Bbb Z$. With this in hand, it is true "by construction" that $\tilde I$ is a homeomorphism, but if you don't quite see that, I encourage you to check this directly by verifying, like you say, that $\tilde I$ and $\tilde I^{-1}$ are continuous. This is "just" a matter of unwinding definitions, but do ask if you get stuck. – 2017-02-15
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    Thanks. I want to verify that $\tilde{I}$ is continuous. Given open subset $U=\{1,2,3\}$ in $\mathbb{Z}$. Then $I^{-1}(U)=[1,4)$. Then $\pi I^{-1}(U)=\tilde{I}^{-1}(U)=\{[1],[2],[3]\}$. However, $\{[1],[2],[3]\}$ is not open in $\mathbb{R}/R$. – 2017-02-15
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    or do you mean open sets on $\mathbb{Z}$ are empty set, $\mathbb{Z}$ and $\{n\in \mathbb{Z}: n\leq N \}$ where $N \in \mathbb{Z}$. – 2017-02-15
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    That's right. We're looking for a topology on $\Bbb Z$ for which $\tilde I$ is a homeomorphism, and our candidate is $\{\emptyset\} \cup \{\{n \in \Bbb Z : n \leq N\} : N \in \Bbb Z\} \cup \{\Bbb Z\}$. In particular, this not does include the subset $\{1, 2, 3\}$. On the other hand, what is, e.g., $\tilde{I}^{-1}(\{n \in \Bbb Z : n \leq N\})$? – 2017-02-16