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.
Best Answer
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.