inclusion $\Bbb R \hookrightarrow \Bbb R^2/\Bbb Z^2$ is an injective morphism but not a homeomorphism I am trying to understand this statement in my course:
inclusion $i:\Bbb R \hookrightarrow \Bbb R^2/\Bbb Z^2$ is an injective morphism but not a homeomorphism.
Before I try to prove it is not a homeomorphism, I don't understand why it is injective:
We have $\Bbb R\cong \Bbb R \times \{1\}$, so for instance :
$i(5) = i(5,1) = 0$ and $i(1) = i(1,1) = 0$.
Am I missing something?
Thank you for your hints and comments.
 A: This statement is not precise, because there is no canonical inclusion of $\mathbb{R}$ in $\mathbb{R}^2/\mathbb{Z}^2$.
However, there are different natural ways to define an application $\mathbb{R} \rightarrow \mathbb{R}^2/\mathbb{Z}^2$.
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1) The first one is, as you do, to define an application $f : \mathbb{R} \rightarrow \mathbb{R}^2/\mathbb{Z}^2$ by
$$f(x) = \overline{(x,1)}$$
(where $\overline{(a,b)}$ denotes here the canonical projection of $(a,b) \in \mathbb{R}^2$ in $\mathbb{R}^2/\mathbb{Z}^2$).
This is obviously not injective because $f(x+1) = \overline{(x+1,1)} = \overline{(x,1)} = f(x)$.
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2) Rather than imbedding $\mathbb{R}$ into $\mathbb{R}^2$ with the map $x \mapsto (x,1)$, you can also see $\mathbb{R}$ as a straight line with a non-zero slope. Consider the application from $\mathbb{R}$ to $\mathbb{R}^2$ defined by $x \mapsto (x,ax)$. Let's see if the induced application $f_a : \mathbb{R} \rightarrow \mathbb{R}^2/\mathbb{Z}^2$ defined by
$$f_a(x) = \overline{(x,ax)}$$
is injective or not.
First you can see that if $a= \frac{p}{q}$ is rational ($p,q \in \mathbb{Z}$), $f_a$ is not injective. Indeed, one has
$$f(0) = \overline{(0,0)} = \overline{(q,p)} = \overline{\left(q, \frac{p}{q} \times q\right)} = f(q)$$
But you can see that is $a$ is irrational, then $f_a$ is injective. Indeed, if $f$ is not injective, then there exists $x \neq y \in \mathbb{R}$ such that $f_a(x)=f_a(y)$, i.e. such that there exists $(p,q) \in \mathbb{Z}^2$ such that 
$$(x,ax) = (y + q,ay + p)$$
You deduce that $x-y = q \neq 0$, and therefore $aq = p$, so $a$ is rational.
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3) Finally, you can prove that in the case where $a$ is irrational (and therefore $f_a$ is injective, $f_a$ is not a homeomorphism. There are several wyas to see that : you can for example see that the image of $\mathbb{R}$ by $f_a$ is not closed (in fact it is dense into $\mathbb{R}^2/\mathbb{Z}^2$). I let you think about this.
