Revisted$_2$: Are doubling and squaring well-defined on $\mathbb{R}/\mathbb{Z}$?

Question

Define a relation $\sim$ on $\mathbb{R}$ as follows: for any $a,b \in \mathbb{R}$, $$a\sim b \iff a-b\in \mathbb{Z}.$$

Let $S=\mathbb{R}/{\sim}$. That is, $S$ is the set of equivalence classes of elements of $\mathbb{R}$ under the equivalence relation $\sim$.

Define $f:S\rightarrow S$ by $f([t])=[t^2])$ for $t\in \mathbb{R}$. Is this well-defined?

Define $g:S\rightarrow S$ by $f([t])=[2t]$ for $t\in \mathbb{R}$. Is this well-defined?

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Let $C=\{(x,y)\in\mathbb{R^2}:x^2+y^2=1\}$, and define $h:S\rightarrow C$ by $$h([t])=(\cos(2\pi t),\sin(2\pi t)).$$

What is the process here to realize whether or not $h$ is well-defined? Or whether it is a bijection?

I know that I have to show $$[t]=[t']\implies h([t])=h([t']),$$ and $$\forall x\in C \exists x'\in S : h([x'])=x,$$ but I'm not really sure how to go about this.

Answer 1

$$t-s\in\mathbb{Z}\implies t^2-s^2=(t-s)(t+s)\in\mathbb{Z}\tag{false}$$

$$t-s\in\mathbb{Z}\implies 2t-2s=2(t-s)\in\mathbb{Z}\tag{true}$$

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$$t-s\in\mathbb{Z}\implies (\cos(2\pi t),\sin(2\pi t))=(\cos(2\pi s),\sin(2\pi s))\tag{true}$$

Answer 2

What you need to check is: suppose $t\sim s$, does it follow that $t^2\sim s^2$ and does it follow that $2t\sim 2s$.

Consider for instance the elements $[\sqrt2]$ and $[1+\sqrt2]$. These elements are the same, since the representatives are equivalent. So, we may write $\alpha=[\sqrt2]=[1+\sqrt2]$. So, if $f$ were well-defined it ought to know what to do with $\alpha $ and produce for us $f(\alpha)$. So, let's see: $f(\alpha)=f([\sqrt2])=[\sqrt2^2]=[2]$ while at the same time $f(\alpha)=f([1+\sqrt 2])=[3+2\sqrt 2]$. Well, this is a problem since $[2]\ne[3+\sqrt2]$ (since the representatives are not equivalent). So $f$ is not well-defined. It does not know what to do with the single element $\alpha$ as input. It can't decide where to send it to just based on a representative. You give it different representatives for the same element and it gives you different elements as answer. Can't do that. You can check that $g$ is well-defined though.