I am studying Hecke operators, and one of the first claims we prove is that there exists a bijection between $\Gamma(1) \backslash \mathfrak{h}$ and $\mathbb{C}^{\times}\backslash \mathcal{L}$. Both the entities indicate a group action. To prove the bijection, we show injectivity and surjectivity. While I understand how to show that a mapping is a bijection, and what group actions are, their combination is for some reason throwing me off.
If someone could explain to me using this example, or another, what exactly we need to show for injectivity and surjectivity and why, I would be grateful. I have expanded a little more on the aforementioned example below:
Let $\mathfrak{h}$ indicate the upper half of the complex plane, and $\Gamma(1)=\operatorname{SL}_2(\mathbb{Z})$. $\Gamma(1)$ acts on $\tau \in \mathfrak{h}$ as $$\begin{bmatrix}a & b\\ c & d\end{bmatrix} \cdot \tau=\frac{a\tau + b }{c\tau +d}.$$ We define $\mathcal{L}:=\{\Lambda \le \mathbb{C} \text{ lattice}\}$ with action by $\mathbb{C}^{\times}$ given by $$z\cdot \Lambda=\{z\lambda : \lambda \in \Lambda\},$$ $ z \in \mathbb{C}^{\times},\Lambda \in \mathcal{L}.$ The map $\tau \mapsto \Lambda_{\tau}:=\mathbb{Z}\tau \oplus \mathbb{Z}$ induces a bijection $$\Gamma(1)\backslash \mathfrak{h} \longleftrightarrow \mathbb{C}^{\times}\backslash \mathcal{L}.$$
Great question. Since you mentioned you're relatively new to group axioms, let's explicitly emphasize that you are not trying to show (and it's not true) the strong statement that $(\mathbb{C}^\times, \mathcal{L})$ and $( \Gamma(1), \mathfrak{h})$ are isomorphic as "sets with groups acting on them." You are only trying to show that the orbit-spaces are in bijection with each other.
This means that given an arbitrary orbit of the action of $\Gamma(1)$ on $\mathfrak{h}$, you need to write down an orbit of the action of $\mathbb{C}^\times$ on $\mathcal{L}$. You can then either figure out an explicit inverse to this map, or show that it's injective and surjective.
Now (as always with quotients) to write down an orbit, we pick a member of it and think of the whole equivalence class.
You mapped $[\tau]$ to $[\Lambda_\tau] := \mathbb{Z} \oplus \mathbb{Z_\tau}$. But this map might not be well-defined -- there is something to check: if $\gamma \tau = \tau'$ for some $\gamma$ in $\Gamma(1)$, you need to be sure that $[\Lambda_\tau] = [\Lambda_{\tau'}]$, i.e., you need to check that that there is some $\lambda \in \mathbb{C}^\times$ with $\Lambda_\tau = \lambda\Lambda_{\tau'}$.
Once you've shown this, you can either show injectivity+surjectivity or try to construct an inverse. Usually you have to do the latter anyway to get to surjectivity, so let's start there.
For surjectivity: given a lattice $\Lambda$, you need to find some $\tau$ and some $\lambda$ such that $\lambda\Lambda = \Lambda_\tau$. You just asked for an explicit statement of what you need to show and not the proof, but as a hint, pick a basis for your lattice, and consider dividing everything by a basis element (you need to be careful, because you need the thing that isn't $1$ to be in the upper half plane).
For injectivity, you need to show that if $\Lambda_\tau$ differs from $\Lambda_{\tau'}$ by a scalar, then $\tau$ differs from $\tau'$ by an element of $\Gamma(1)$. Note that this is the same as saying the inverse "map" alluded to in the surjectivity paragraph above is actually well-defined. As a hint, consider, once $\tau$ is fixed, what other numbers $\tau'$ are such that the lattices $\Lambda_\tau$ and $\Lambda_{\tau'}$ are literally equal, not just equivalent.
Again, this rather lengthy answer is just enumerating what you're supposed to check, and not giving a proof. Giving a proof can be found in many places if you are still stuck!