Consider $\mathbb{Z}$ with the digital line topology, which has as a basis the sets $\{n\}$ for $n$ odd, and $\{n-1,n,n+1\}$ for $n$ even, and consider the partition of $\mathbb{Z}$ created using modulo $10$ arithmetic. The partition gives the mapping:
$$\{\ldots, –30, –20, –10, 0, 10, 20, \ldots\} \to 0$$
$$\{\ldots, –29, –19, –9, 1, 11, 21, \ldots\} \to 1$$
$$\{\ldots, –28, –18, –8, 2, 12, 22, \ldots\} \to 2$$ $$ \cdots$$ $$\{\ldots, –21, –11, –1, 9, 19, 29, \ldots\} \to 9$$
How do I know which sets are open in quotient topology on $\{0,1,2\ldots,9\}$?
Here are some hints/comments.
The digital line topology on $\mathbb{Z}$ has as a basis the sets $\{n\}$ for $n$ odd, and $\{n-1,n,n+1\}$ for $n$ even.
Consider the quotient map $q\colon\mathbb{Z}\to\mathbb{Z}/10\mathbb{Z}$. Then $U\subseteq\mathbb{Z}/10\mathbb{Z}$ is open, iff $q^{-1}(U)$ is open in $\mathbb{Z}$ in the digital line topology.
Let's understand this property. If $0<n<10$ is odd, and I interpret it as an element in the quotient space, then $q^{-1}(\{n\}) = \{n+10r\ |\ r\in\mathbb{Z}\}$, i.e. the preimage consists of a disjoint collection of odd numbers, which is open.
If $n$ had been even, you see that the preimage would not have been open.
Can the preimage of a set of two numbers $\{n,m\}$, where they are not both odd, ever be open?
What if you consider the preimage $q^{-1}(\{0,1,2\})$? How about $q^{-1}(\{1,2,3\})$?
When is the preimage of three consecutive integers open?