Let $\mathbb{T}^2=[0,1) \times [0,1)$ be the $2$-torus. Let $\lambda$ be an irrational no. Prove that the map $\gamma:\mathbb{R} \to \mathbb{T}^2$ defined by $\gamma(t)=(t,\lambda t)(\mod 1)$ is an injective immersion but not an embedding.
$\gamma$ is one-to-one: If $\gamma(t_1)=\gamma(t_2)$, then $(t_1,\lambda t_1)(\mod 1)=(t_2,\lambda t_2)(\mod1)$. This means that $t_1-t_2 \in \mathbb{Z}$ and so is $\lambda(t_1-t_2)$. If $t_1-t_2\ne0$, then we have that $\lambda=\frac{p}{q} \in \mathbb{Q}$ which is a contradiction.
If I look at $\gamma'(t)$, then atleast one of the coordinates is always $1$. This is because, the derivative of a fractional part is $0$ at an integer point and since $\lambda$ is an irrational, both coordinates can't be an integer except when $t=0$. So, $\gamma'(t)$ doesn't vanish except when $t=0$. How do I conclude from here that $\gamma$ is an immersion?
By Dirichlet's Approximation Theorem, given $\lambda \in \mathbb{R}$ and any positive integer $N$, there exists $n,m$ with $1 \le n \le N$ such that $|n\lambda-m| \lt \frac{1}{N}$. Then $|\gamma(n)-\gamma(o)|=|(0,\{\lambda n\})-(0,0)|=|(0,\{\lambda n\})-(0,\{m\})|=|\{\lambda n\}-\{m\}|=|\{\lambda n-m\}| \le |n\lambda-m| \lt \frac{1}{N}$
This shows that $\gamma(0)$ is a limit point of $\{\gamma(\mathbb{Z})\}$. This means that $\gamma$ is not a homeomorphism as $\mathbb{Z}$ has no limit points in $\mathbb{R}$. Thus,$\gamma$ is not an embedding.
Is this alright?
Thanks for the help!!