I've been told to identify the terms in the canonical decomposition of the function r |-> exp(2*pi*i*r) from R -> C.

I've been able to give an answer, but I think i might have misinterpreted the question, because it doesn't feel like i did anything. This is my answer:


A is R (reals)

Z is C (complex)

let f: R -> C be r |-> exp(2*pi*i*r)

Pick = as the equivalence relation ~

then A/~ is the set of singletons {{x} | x in R}

the function q: A -> A/~ can be written as q(x) = {x}

by definition, f~([a]~) = f(a) where f~ is the function from A/~ to im f

we know [a]~ = {a}, therefore, f~([a]~) = f~({a}) = f(a) = exp(2*pi*i*a),

f~({a}) = exp(2*pi*i*r)

the third function is the inclusion v: im f -> B, v(x) = x


The reason i think I'm missing something is because i could have replaced the function r |-> exp(2*pi*i*r) with pretty much anything else, and just substituted that expression in two places without changing the answer. Am I missing something?

I've searched around for more about 'canonical decomposition', but it doesn't seem to be a widely used term at all.


1 Answer 1


If you read your link, you can see that the relation is not chosen, she is determined by the choice of $f$ :

$x\sim{} y\Leftrightarrow f(x)=f(y)$.

So in this case, you can show that the relation is $x\sim{}y\Leftrightarrow e^{2i\pi x}=e^{2i\pi y}\Leftrightarrow x\equiv y \mod 1$

Then your quotient is $\mathbb{R}/{\sim{}}=\{x+\mathbb Z : x\in \mathbb R\}$ where $x+\mathbb Z= \{x+n: n\in\mathbb Z\}$.

So $q(x)=x+\mathbb{Z}$ and $\tilde{f}(x+\mathbb{Z})=exp(2i\pi x)$ (v is like in your question).


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