Homework Help: Topology question concerning global continuity of the canonical map.

1. Jun 13, 2012

Wodfrag

1. The problem statement, all variables and given/known data
If the set \Z of integers is equipped with the relative topology inherited from ℝ, and κ:\Z→\Z_n (where κ is a canonical map and \Z_n is the residue class modulo n) what topology/topologies on \Z_n will render κ globally continuous?

2. Relevant equations

3. The attempt at a solution
I think i have found that κ will be globally continuous if \Z_n is equipped with the trivial (where the open sets are ∅ and \Z_n itself) topology, since \Z_n itself will be an open neighborhood of an element in \Z_n, hence for every x \in \Z κ(x)\in \Z_n. I cannot figure out however if i can apply any other topologies on \Z_n such that κ will be globally continuous.

Thanks

2. Jun 14, 2012

algebrat

Since intuition may help to decide which way to go, try some of these tricks that I love, but sadly rarely come in handy:(

In order for the map to remain continuous, close points must map to close points. And the more open sets you have, the farther apart points are. But in the discrete topology, every set is an open set, you have as many open sets as possible, so points are really far apart.

That means that no points are close together.

So when I said that close points map to close points, we have no close points, so they can map as far apart as we like. Okay the logic is getting weird, but here's where I'm thinking we are at this point:

We can have any topology in Z_n. Then any open set will pull back to an open set, because every set in Z is open.

Perhaps every map out of the discrete topology is continuous.

Last edited: Jun 14, 2012
3. Jun 14, 2012

HallsofIvy

The set of integers, Z, is discrete and the topology inherited from R is the discrete topology. All sets are open (and closed) so that any function from Z to any topological space is continuous.

4. Jun 14, 2012

Wodfrag

Thanks for the answers! You make it sound so very simple..

5. Jun 14, 2012

Wodfrag

Thanks for the answers! You make it sound so very simple..