Linear transformation and restriction map 
Let $T: V \to W$ be a linear transformation between 2 vector spaces
  over a field $F$. Show that there exists a vector subspace $V_1$ of
  $V$ such that the restriction map $$T_1 = T|V_1 : V_1 \to W$$ $$v_1
 \mapsto T(v_1)$$ satisfies the two conditions below:
i) $T_1$ is a linear injective mapping.
ii) Image of $T_1$ is equals to Image of $T$ 

Do note we cannot assume that $V$ and $W$ are finite dimensional.
This is an past year exam question which i had no idea how to approach.
My first attempt is as follows:
Let the basis of $V_1$ be $B_1 = \{v_i~|~ i \in A_1\}$ and basis of $W$ be $C = \{w_i ~|~ i \in A_2\}$ where $A_i$ is a index set which can be infinite.
So i try to mimic the "approach when $V$ and $W$ are finite dimensional" and claim that the mapping $T_1$ maps $\sum_{i=1} a_iv_i$ to $\sum_{i=1}b_iw_i$.
I am not sure how to continue, can anyone help me?
 A: Knowing absolutely nothing about the vector spaces $V$ and $W$, one is forced to then use the structural fact that every vector space has a basis (a maximal linearly independent set) using Zorn's lemma.
Then, let $\mathcal W = \{w_i\}$ be a basis for the image of $T$, which we shall call as $Z$. Let $v_{w_i}$ be any preimage of $w_i$, for each $i$.
Now, consider the span of $\{v_{w_i}\}$, which is a subset of $V$. Call this $V_1$.
I claim that $T$ restricted to $V_1$ does the job.
We will first show that the image of $T|_{V_1}$ equals $Z$. It clearly is contained in $Z$. However, every $z \in Z$ can be written as a linear combination $z = \sum z_i w_i$, so $z = T(\sum z_iv_{w_i}) \in T(V_1)$, proving the other containment.
Suppose  that $T(v) = 0$. Note that $v \in V_1$, so it can be written as a linear combination $v = \sum c_iv_{w_i}$. Then, $T(v) = \sum c_iw_i = 0$, but since the $w_i$ are linearly independent, this implies $c_i = 0$ for all $i$, and hence $v = 0$. Therefore, $T|_{V_1}$ is injective. 
