Assume for all subsets $S\subseteq V$ that if $T(S)$ spans $W$, then $S$ spans $V$. Prove that $T$ is injective. Question: Suppose $V,W$ are vector spaces and let $T:V\rightarrow W$ be a surjective linear transformation.  Assume for all subsets $S\subseteq V$ that if $T(S)$ spans $W$, then $S$ spans $V$.  Prove that $T$ is injective.
My thoughts: Since $T$ is surjective, we can find $v\in V$ such that $T(v)=w$ for any $w\in W$.  We are given that if $T(S)$ spans $W$ then $S$ spans $V$, so we can write $w=T(v)=\sum_n \alpha_n T(s_k)$, where $\alpha_i$ are scalers and $s_i\in S$.  Now, if we let $v_1,v_2\in V$, to show that $V$ is injective, I need to show that if $T(v_1)=T(v_2)$ then $v_1=v_2$ (or the equivalent ``not equal" statement).  But, since we wrote $T(v)$ as a sum that doesn't depend on $v$, wouldn't it be "obvious" (what a dangerous word to use here since I am not sure anyway).  Any help, insight, or assistance is greatly appreciated!  Thank you.
 A: Take a basis $B \subseteq W$. For each $b \in B$, pick $s_b \in V$ such that $T(s_b) = b$.
Then $S = \{s_b \mid b \in B\}$ must span all of $V$, since $T(S) = \{T(s_b) \mid b \in B\} = B$ spans all of $W$.
Now suppose we have $T(x) = 0$. Since $S$ spans $T$, write $x = \sum\limits_{i = 1}^n c_i s_{b_i}$, where $c_i$ is a scalar and $b_i \in B$ for all $i$. Then we have $T(x) = 0 = \sum\limits_{i = 1}^n c_i b_i$.
Since $B$ is a basis, we see that for all $i$, we must have $c_i = 0$. Therefore, $x = 0$.
Thus, $T$ is injective.
I don't see a way to do this without invoking the axiom of choice at some point, which is rather unfortunate.
A: Let $x\in V$, $x\ne0$. Then $\{x\}$ is a linearly independent set, so there exists a basis $\mathscr{B}$ of $V$ such that $x\in\mathscr{B}$ (see final note). Then, by surjectivity, $T(\mathscr{B})$ spans $W$.
If $T(x)=0$, then also $T(\mathscr{B}\setminus\{x\})$ spans $V$, so $\mathscr{B}\setminus\{x\}$ spans $V$: contradiction.
Therefore $T(x)\ne0$.
Final note.  I don't think you can avoid existence of bases, hence the axiom of choice, if you want the result for arbitrary vector spaces; no problem for finite dimensional spaces.
A: Let $v \neq 0$ be such that $T(v)=0$. Extend this to a basis of $V$. Let's call that basis $B$. Now as $T$ is surjective $T(B) $ spans $W$. But $T(v) =0$ so $T(B \setminus v) $ spans $W$ which means $B\setminus v$ spans $V$ a contradiction as $B$ is a basis. Thus our assumption $v\neq 0$ is false.
