Prove that if ${(v_1- v_2) } \in U$ where $U$ is a subspace of $V$ then $v_1 + U$ = $v_2 + U$ where $v_1 , v_2$ belong in $V$

Question

I assumed $v_1 - v_2$ to be equal to some $u \in U$.

Then I wrote $u$ = $u_2 - u_1$.

So now $v_1 - v_2 = u_2 - u_1$ for any $u_2,u_1 \in U$.

Therefore $v_1 + u_1= v_2 + u_2$ for any $u_2,u_1 \in U$

Thus $v_1 + U_1= v_2 + U_2$

I still got this wrong on my test.

The correct solution was to write

$v_1 + u= v_2 + (v_1 - v_2) + u$. Now since $(v_1 - v_2) \in U$

$(v_1+u) \in (v_2 + U)$. Now you can go in the other direction and prove that $(v_2+u) \in (v_2+ U)$.

Thus $v_1 + U$ = $v_2 + U$.

Now I understand that the second proof look more formal but I don't know why my proof seems incomplete according to my professor

Answer 1

Can we do it this way?

Let's consider a map $T$ $\in$ $\mathcal{L}(V, V/U)$. Where the quotient space is defined as $$ V/U := \{ v + u~:~\text{for any } v \in V \text{ and } u \in U, \text{ where } U \subseteq V \} $$ We know that the map is defined as
\begin{align*} T :&~ V \to V/U\\ & v \mapsto v + U \end{align*} And we know that $\text{ker}(T) = U$. According to the question $v_{1} - v_{2}$ $\in$ $U = \text{ker}(T)$, where $v_{1}, v_{2}$ $\in$ $V$. Therefore we have \begin{align*} &T(v_{1} - v_{2}) = 0\\ \implies & T(v_{1}) - T(v_{2}) = 0 \quad [\text{as } T \in \mathcal{L}(V, V/U)]\\ \implies & T(v_{1}) = T(v_{2})\\ \implies & v_{1} + U = v_{2} + U \end{align*}

Is this approach wrong?