# Group of homomorphisms from $\mathbb{R}/\mathbb{Z}$ to itself

Consider the (additive) group $$G$$ of all homomorphisms $$\phi: \mathbb{R}/\mathbb{Z} \rightarrow \mathbb{R}/\mathbb{Z}$$.

I think there is an injective homomorphism $$\mathbb{Z} \rightarrow G$$ where an integer $$q$$ is sent to the map $$[x] \mapsto [qx]$$ for $$[x] \in \mathbb{R}/\mathbb{Z}$$.

I'm wondering if this injection is also a surjection? In other words are there endomorphisms of $$\mathbb{R}/\mathbb{Z}$$ that do not come from integers?

If you allow $$\phi$$ discontinuous (and assume the axiom of choice), there are such "exceptional" endomorphisms. For example, given a Hamel basis with generating set $$B$$ and $$1\in B$$, designate any $$1\neq b\in B$$ and map all elements of $$B$$ besides $$b$$ to $$0$$.

On the other hand, if you wish only to consider continuous endomorphisms, then you do have all of them. To show this, take a continuous $$\phi:\mathbb R/\mathbb Z\to\mathbb R/\mathbb Z$$, and take its pullback $$\pi:\mathbb R\to\mathbb R/\mathbb Z$$. Identify $$\mathbb R/\mathbb Z$$ with $$[-1/2,1/2)$$, and consider some $$\epsilon>0$$ for which $$\pi((-\epsilon,\epsilon))\subset (-1/4,1/4)$$. Now, consider a map $$\tilde\pi:\mathbb R\to\mathbb R$$ constructed by extending $$\pi$$ on $$(-\epsilon,\epsilon)$$ to all of $$\mathbb R$$ by $$\tilde\pi(2x)=2\tilde\pi(x)$$. This map is continuous since, for any $$x$$, there exists some $$n$$ for which $$\tilde\pi(y)=2^n\pi(y/2^n)$$ holds for all $$y$$ near $$x$$ (with $$|y/2^n|<\epsilon$$). It's clearly additive for similar reasons (dividing by a large power of $$2$$). So, $$\tilde\pi:\mathbb R\to\mathbb R$$ is a continuous homomorphism.

We claim that this implies $$\tilde\pi(x)=\alpha x$$ for some fixed constant $$x$$. If not, take some distinct $$x$$ and $$y$$, linearly independent over $$\mathbb Q$$, for which $$\tilde\pi(x)=\alpha x$$ and $$\tilde\pi(y)=\beta y$$ and $$\alpha\neq \beta$$. Since $$\mathbb Zx+\mathbb Zy$$ contains nonzero reals arbitrarily close to $$0$$, but $$\alpha mx+\beta ny$$ doesn't tend to $$0$$ as $$mx+ny\to 0$$ and $$(m,n)$$ grows (it looks like $$(\beta-\alpha)ny$$), this contradicts continuity at $$0$$.

So, $$\tilde\pi(x)=\alpha x$$ for some real $$\alpha$$, which implies that $$\pi(x)=\alpha x$$ for some real $$\alpha$$. The fact that $$\pi(1)=0$$ implies that $$\alpha$$ is an integer, which finishes the proof.

Yes, if you require the homomorphisms to be continuous (see the other answer otherwise).

Note for any (locally compact, abelian) group $$G$$ the group of continuous homomorphisms $$\{ \phi : G \to \mathbb{R} / \mathbb{Z} \}$$ is the Pontryagin Dual $$G$$, often written $$\widehat{G}$$.

Then you're asking for the pontryagin dual of the circle group, but this is well known to be $$\mathbb{Z}$$. Moreover, if you chase through the definitions, you'll find these characters are exactly the same copy of $$\mathbb{Z}$$ that you've found (though they're traditionally denoted $$z \mapsto z^n$$ instead of $$[x] \mapsto [nx]$$ since in this context we usually view the circle group as the unit circle in $$\mathbb{C}$$ under multiplication).

I hope this helps ^_^