# Is the image of this strip equal to the modular curve $Y(2)$

Let $B$ be the image of the strip $$\{x+iy : -1\leq x< 1, y>\frac{1}{2} \} \subset \mathbf{H}$$ in the modular curve $Y(2)$ under the quotient map $\mathbf{H} \to Y(2)$.

Is $B=Y(2)$?

Is $B=Y(2)$ iff the strip $\{x+iy : -1\leq x< 1, y>\frac{1}{2}\}$ contains a fundamental domain for $\Gamma(2)$?

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No, your strip does not surject onto $Y(2)$.
Domains $A \subseteq \mathbf{H}$ have a $\operatorname{PSL}(2,\mathbb{R})$-invariant "area" given by
$$\int_{A} \frac{\mathrm{d}x\, \mathrm{d}y}{y^2}$$
Your strip has area 4, but the standard fundamental domain for $\operatorname{PSL}(2, \mathbb{Z})$ has area $\frac{\pi}{3}$. As $\Gamma(2)$ has index 6 in $\operatorname{PSL}(2,\mathbb{Z})$ and $6 \times \frac{\pi}{3} > 4$, there's no way it can surject onto $Y(2)$.
(PS: Here's another reason: $\Gamma(2)$ has 3 cusps, and your set contains a neighbourhood of $\infty$ but not of any other cusp. So there's no way that any set contained in $\{ z: \operatorname{Im}(z) > r\}$, for any $r > 0$, can surject onto $Y(2)$.)