Homomorphisms of the group of affine trasformations I'm trying to prove the following theorem.

Let $\mathrm{Aff}(n)$ define the set of transformations $T: \mathbb{R}^n \to \mathbb{R}^n$ sending $x \mapsto Ax + b$ such that $A \in \mathrm{GL}_n (\mathbb{R})$ and $b \in \mathbb{R}^n$. Prove that (a) $\psi: \mathrm{Aff}(n) \to \mathrm{GL}_n (\mathbb{R})$ sending $T \mapsto A$ is a homomorphism of groups, but that (b) $\phi: \mathrm{Aff}(n) \to \mathbb{R}^n$ sending $T \mapsto b$ is not.

My key problem in (a) is formalism. I want to act on an arbitrary element of $x$, but I can't do that because inputs to $\psi$ are elements of $\mathrm{Aff}(n)$, not of $\mathbb{R}^n$, so I need to do the casework "outside" of the main equation chain, since it doesnt make sense for me to write $\psi((T_1 \circ T_2)(x))$ or $\varphi(T_1)(x)$. With that said, here is my attempt that I believe to be insufficiently formal.

Let $T_i \in \mathrm{Aff}(n)$ sending $x \longmapsto A_i x + b_i$ for $1 \leq i \leq n$. For any $x \in \mathbb{R}^n$, we h ave
\begin{align*}
(T_1 \circ T_2)(x) & = T_1(T_2 (x)) \\
& = T_1 (A_2 x + b_2) \\
& = A_1 (A_2 x + b_2) + b_1 \\
& = (A_1 A_2)x + (A_1 b_2 + b_1)
\end{align*}
So
\begin{align*}
\psi(T_1 T_2) = A_1 A_2 = \psi(T_1) \psi(T_2).
\end{align*}

The problem, again, is that I can't define the output of $\psi$ without referring to $x$.

I'm less sure on part (b), so this is more scratchwork than much else. Take $T_1, T_2$ defined as above. Then, using the above work, we get:
\begin{align*}
\phi(T_1 T_2) & = A_1 b_2 + b_1 \\
\phi(T_1) + \phi(T_2) & = b_1 + b_2
\end{align*}
But $A_1 b_2 + b_1 \neq b_1 + b_2$ in general. Indeed, we have:
\begin{align*}
A_1 b_2 + b_1 = b_1 + b_2 & \iff A_1 b_2 = b_2 \iff A_1 = I.
\end{align*}
So specifying $T_1$ with any $A_1 \neq I$ gives a counterexample.

This is not an explicit counterexample and I don't quite know how to construct one for every $n \in \mathbb{N}$.
 A: If I understand you correctly, you struggle to show that $\psi$ satifsies the homomorphism property
$$\phi (T \circ S) =\phi(T) \phi(S)$$
Suppose $T(x) = Ax + a$ and $S(x) = Bx + b$. We have
$$(T \circ S)(x) = T(Bx + b) = A(Bx+b)+a = (AB)x + (Ab + a)$$
Hence, $\psi(T \circ S) = A B = \psi(T) \psi(S)$.
Now, you see from the computation above why $\phi$ is not a homomorphism. Since
$$\phi(T \circ S) = Ab + a,$$
which in general does not equal $\phi(T) + \phi(S) = a + b$.
We give the following counter example: Let $T$ and $S$ be given by
\begin{align}
T(x) &= \begin{bmatrix} 
0 & 0 \\
1 & 0 \end{bmatrix} x + \begin{bmatrix}
1 \\
0 \end{bmatrix} \\
S(x) &= \begin{bmatrix}
1 & 0 \\
0 & 0 \end{bmatrix} x + \begin{bmatrix} 
1 \\
2 \end{bmatrix}
\end{align}
A simple computation shows that
$$(T \circ S)(x) = \begin{bmatrix} 
0 & 0 \\
1 & 0 \end{bmatrix}x + \begin{bmatrix} 
1 \\
1 \end{bmatrix}$$.
Hence, $\phi(T \circ S) = (1,1)^\top$, whereas $\phi(T) + \phi(S) = (1,0)^\top + (1,2)^\top = (2,2)^\top$.
