From S.L Linear Algebra:
Let $V$ be a vector space, and let $u$ be a fixed element of $V$. We let $T_u: V \rightarrow V$ be a map such that $T_{u}(v)=u+v$. We call $T_{u}$ a translation by $u$.
If $S$ is any subset of $V$, then $T{u}(S)$ is called translation of $S$ by $u$, and consists of all vectors $u$+$v$, with $v \in S$. We often denote it by $S+u$. In the picture below, we draw a set $S$ and its translation by a vector $u$.
Prove that if $u_1$, $u_2$ are elements of $V$, then $T_{u_1+u_2}:T_{u_1} \circ T_{u_2}$. Also, prove that if $u$ is an element of $V$, then $T_{u}:V \rightarrow V$ has an inverse mapping which is nothing but translation $T_{-u}$.
Note:
"$\circ$" symbol in the text above denotes composite mapping.
Solutions
Axiom N1: Prove that if $u_1$, $u_2$ are elements of $V$, then $T_{u_1+u_2}:T_{u_1} \circ T_{u_2}$.
Proof:
Let $S$ be a subset of $V$, then $T_{u_1+u_2}(S)=\{v+(u_{1}+u_{2}) \mid v \in S \}$ which is equal to $T_{u_1} \circ T_{u_2}(S)=T_{u_1}(T_{u_2}(S))=\{(v+u_{2})+u_{1} \mid v \in S\}$ due to the commutativity and associativity axioms declared in vector spaces.
Axiom N2: Also, prove that if $u$ is an element of $V$, then $T_{u}:V \rightarrow V$ has an inverse mapping which is nothing but translation $T_{-u}$.
Proof:
Let $S$ be a subset of $V$, then $T_u(S)=\{v+u \mid v \in S\}$. Inverse mapping of $T_{u}(S)$ then must be a function $L$ such that $L_{t}(T_{u}(S))=I=0$ where $I$ is additive identity. This is only possible if $t=-u$.
Are both proofs valid? I'm slightly worried that they might have few mistakes in them. Perhaps there are even better ways to prove it?
Thank you!
You are supposed to work with elements of $V$, instead of subsets. Asserting that $T_{u_1+u_2}=T_{u_1}\circ T_{u_2}$ means that, for each $v\in V$,$$T_{u_1+u_2}(v)=T_1\bigl(T_{u_2}(v)\bigr),$$which is easy:\begin{align}T_{u_1+u_2}(v)&=(u_1+u_2)+v\\&=u_1+(u_2+v)\\&=T_1\bigl(T_{u_2}(v)\bigr).\end{align}The same objection applies to the other proof.