Let $V$ and $W$ be vector spaces with subspaces $V_{1}$ and $W_{1}$, respectively. If $T:V\to W$ is linear, prove that $T(V_{1})$ is a subspace of $W$ and that $\{x\in V: T(x)\in W_{1}\}$ is a subspace of $V$.
MY ATTEMPT
Since $0_{V}\in V_{1}$, $T(0_{V}) = 0_{W}\in T(V_{1})$.
Let us suppose that $a\in T(V_{1})$ and $b\in T(V_{1})$. We have to prove that $a + b \in T(V_{1})$ as well as $\alpha a\in T(V_{1})$ where $\alpha\in\textbf{F}$.
Indeed, as a consequence of forward image definition, there are $x\in V_{1}$ and $y\in V_{1}$ such that $a = T(x)$ and $b = T(y)$.
Hence $a + b = T(x) + T(y) = T(x+y)\in T(V_{1})$.
Similarly, $\alpha a = \alpha T(x) = T(\alpha x)\in T(V_{1})$, and we are done with the first part.
Let us solve the second part now.
Indeed, $0_{V}\in\{x\in V: T(x)\in W_{1}\}$. This is because $T(0_{V}) = 0_{W}\in W_{1}$.
If $c,d\in\{x\in V:T(x)\in W_{1}\}$, then $c+d\in\{x\in V:T(x)\in W_{1}\}$ and $\beta c\in\{x\in V:T(x)\in W_{1}\}$.
This is because $T(c+d) = T(c) + T(d)$, $T(\beta c) = \beta T(c)$ and $W_{1}$ is a vector space.
Any comments on my solution? Could someone suggest improvements?
A little (nothing essentially, is more like the presentation) improve:
First, $0_W = T(0_V) \in T(V_1)$ because $0_V \in V_1$. Now, let $w_1$ and $w_2$ in $T(V_1)$. This means that there is vectors $v_1$ and $v_2$ in $V_1$ such that $w_1 = T(v_1)$ and $w_2 = T(v_1)$; next, since $V_1$ is a subspace of $V$, for any $\alpha \in \mathbf F$ we have $\alpha v_1 + v_2 \in V_1$, and then $$\alpha w_1 + w_2 = \alpha T(v_1) + T(v_2) = T(\alpha v_1 + v_2) \in T(V_1).$$
Observe that $\{x \in V :\, T(x) \in W_1\} = T^{-1}(W_1)$. Right, the zero vector $0_V$ is in $T^{-1}(W_1)$ because $T(0_V) = 0_W \in W_1$. Now, let $v_1$ and $v_2$ two vectors in $T^{-1}(W_1)$, this means that $T(v_1)$ and $T(v_2)$ are in $W_1$, and since $W_1$ is a subspace of $W$, for any $\alpha \in \mathbf F$ we have $T(\alpha v_1 + v_2) = \alpha T(v_1) + T(v_2) \in W_1$, that is, $\alpha v_1 + v_2 \in T^{-1}(W_1)$.