Showing a subset is a submanifold of $T^2$ Let $Z=\{[x,y]\in T^2:y = \frac{1}{3} \sin{2 \pi x}+m,m\in \mathbb{Z}\}$. 
I have to show this is a submanifold of $T^2$. So I think the way to proceed is to show that for any $z\in Z$ there is a chart $(U,\phi)$ on $T^2$ such that $\phi(U \cap Z) = \mathbb{R} \times \{0\}$. So this is what I've tried ($q$ will denote the quotient map):
First I write $Z$ as (thinking of $T^2$ as a quotient space and choosing $m$ so that the second coordinate lies in the unit square):
$Z = \{(x,\frac{1}{3}\sin{2 \pi x}):x\in [0,1/2]\} \cup \{(x,\frac{1}{3}\sin{2 \pi x}+1):x\in (1/2,1)\}. $
So my idea is for $x\in (0,1/2)$ choose the coordinate chart: $(U_1,h_1)$, where:


*

*$U_1 = q(0,1/2) \times q(0,1/2)$.

*$h_1([x,y])=(x,y-\frac{1}{3}\sin{2 \pi x})$. Clearly $h_1(U_1\cap Z)=\mathbb{R}\times \{0\}$.


Similarly for $x\in (1/2,1)$ I would take $(U_2,h_2)$ with:


*

*$U_2 = q(1/2,1) \times q(-1/2,0)$.

*$h_2([x,y])=(x,y-\frac{1}{3}\sin{2 \pi x}-1)$. 


Obtaining again $h_2(U_2\cup Z) = \mathbb{R} \times \{0\}.$
For the 2 missing points $(0,0)$ and $(1/2,0)$ I would just take small neighbourhoods and $h_1$ again. Is this approach correct? Any corrections or remarks are very welcome.
 A: I haven't really read your entire solution, but the approach of the adapted charts is at least theoretically correct. One thing I have to point out, though: $\phi(U\cap Z)$ may not be $\mathbb{R}\times \{0\}$, but the definition does not require that, it's enough that $\phi(U\cap Z)=V\times \{0\}$, where $V\subset\mathbb{R}$ is open.
Now, since your $Z$ has the property that whenever $[x,y]\in Z$, $[x+n,y], [x,y+m]$ also lie in $Z$ for any $n,m\in \mathbb{Z}$, I would try a different approach: another way to prove something is a submanifold, is to prove it is preimage of a regular value for a smooth map from your ambient space to another manifold.
Define $f:\mathbb{R}^2\rightarrow \mathbb{R}$ as $f(x,y)=\sin\big(\pi(y-\frac{1}{3}\sin(2\pi x))\big)$. Observe that $f(x+n,y)=f(x,y)=f(x,y+m)$ for any $n,m\in \mathbb{Z}$. So, $f$ induces to a function $\hat{f}:\mathbb{T}^2=\frac{\mathbb{R}^2}{\mathbb{Z}^2}\rightarrow \mathbb{R}$. Since $f$ is clearly $C^{\infty}$ and the quotient map $\pi: \mathbb{R}^2\rightarrow \mathbb{T}^2$ is a local diffeomorphism, $\hat{f}$ is also $C^{\infty}$.
Moreover, $Z=\hat{f}^{-1}(0)$, so it's enough to prove that $0$ is a regular value of $\hat{f}$. Again since $\pi$ is a local diffeomorphism and $f=\hat{f}\circ \pi$, this is equivalent to $0$ being a regular value of $f$.
Observe that $\frac{\partial{}f}{\partial y}=\pi \cos\big(\pi(y-\frac{1}{3}\sin(2\pi x))\big)=\pm\pi$, for points in $f^{-1}(0)$. So that $\nabla f\neq 0$ in $f^{-1}(0)$, and $0$ is a regular value of $f$, hence of $\hat{f}$.
