Fix an integer $n\geq 0$, and let $V_n$ be the complex vector space of polynomials in two variables $z_1$ and $z_2$ homogeneous of degree $n$. Define a representation $$\phi_n:SL(2,\mathbb{C})\to GL(V_n)$$by $$\phi_n(A)P\left(\begin{bmatrix} z_1 \\ z_2 \end{bmatrix}\right)=P\left(A^{-1}\begin{bmatrix} z_1 \\ z_2 \end{bmatrix}\right)$$
I want to show that it is really a representation. For any $P\left(\begin{bmatrix} z_1 \\ z_2 \end{bmatrix}\right)\in V_n$ and any $A,B\in SL(2,\mathbb{C})$, we have \begin{align} \phi_n(AB)P\left(\begin{bmatrix} z_1 \\ z_2 \end{bmatrix}\right)=P\left((AB)^{-1}\begin{bmatrix} z_1 \\ z_2 \end{bmatrix}\right) \end{align} On the other hand $$\phi_n(A)\phi_n(B)P\left(\begin{bmatrix} z_1 \\ z_2 \end{bmatrix}\right)=\phi_n(A)P\left(B^{-1}\begin{bmatrix} z_1 \\ z_2 \end{bmatrix}\right)=P\left(A^{-1}B^{-1}\begin{bmatrix} z_1 \\ z_2 \end{bmatrix}\right)$$ But we do not get $$\phi_n(AB)=\phi_n(A)\phi_n(B)$$
What is wrong here? Thanks!
You are misinterpreting the order of operations. In general, if $G$ acts on a set $X$, then you automatically get an action on the space $Fun(X,Y)$ of functions $\phi:X\to Y$ ($Y$ being any set) by the formula $$(g.\phi)(x)=f(g^{-1}.x).$$ Now, when you write $g.(h.\phi)$, this is read as $g$ acting on the function $(h.\phi)$. According to our formula this should be evaluated as \begin{align*} (g.(h.\phi))(x)&=(h.\phi)(g^{-1}.x)\\ &=\phi(h^{-1}.(g^{-1}.x))\\ &=\phi((h^{-1}g^{-1}).x)\\ &=\phi((gh)^{-1}.x)\\ &=((gh).\phi)(x) \end{align*}
In this specific case, you have the natural action of $SL(2,\mathbb{C})$ on $\mathbb{C}^2$ which yields an action on $Fun(\mathbb{C}^2,\mathbb{C})$. Now, this action restricts to the space $V=Poly(\mathbb{C}^2,\mathbb{C})$ of polynomial functions and again to $V_n$ of homogeneous polynomial functions of degree $n$.