Suppose I am given $G_1,H_1$ as groups and $f_1: G_1 \to H_1 $ a group homomorphism. Then $$\ker f_1 := \{g_1 \in G_1 : f_1(g_1) = id_{H_1}\} \tag{1}$$
Suppose I am given $G_2, H_2$ as vector spaces and $f_2: G_2 \to H_2 $ a linear transformation. Then $$\ker f_2 := \{\mathbf{g_2} \in G_2 : F_2(\mathbf{g_2}) = 0_{H_2}\} \tag{2} $$
If $H_2$ is a vector space, then $id_{H_2}$ is the identity linear transformation. But $ 0_{H_2} $ is $ 0(\mathbf{h_2}) = \mathbf{0} \text{ } \forall \mathbf{h_2}\in H_2. $ so it's different! Shouldn't $0_{H_2}$ in (2) really be $id_{H_2}$ instead?
Edit 2 hours later: I'm working with Möbius maps and the textbook says
"The kernel of the homomorphism $\phi: GL(2, \mathbb{C}) \to \text{(Möbius group)} $ via $$ \phi: \left[ \begin{matrix} a & b \\ c & d \\ \end{matrix} \right] \to f(z) = \dfrac{az + b}{cz + d} \in \mathbb{C} $$ is the set of matrices $\mathbf{A}$ such that the Möbius map $\phi(\mathbf{A})$ is the identity map. So to find the kernel, solve for $z$ in $f(z) = \dfrac{az + b}{cz + d} = z $.
Thanks to all of you, I now understand that $id_{H_1}$ in (1) means the identity element in $H_1$, NOT identity map. But why does the textbook say "map" then? Shouldn't it say "element"? I understand the identity element in the Möbius group is $f(z) = z$.