I have a scenario, and I am wondering if property $(2)$ below is necessarily true. Is it true?
Let $V$ and $W$ be non-zero vector spaces, and $T$ be a surjective linear transformation of $V$ onto $W$. Assume that property $(1)$ holds:
for any subset $S$ of $V$, we have that "$TS$ spans $W$" implies "$S$ spans $V$".
Note that because $T$ is onto, the converse $(1)^C$ holds:
for any subset $S$ of $V$, we have that "$S$ spans $V$" implies "$TS$ spans $W$".
Now suppose we also have a finite-dimensional non-zero subspace $Z$ of $V$, and its non-zero image $TZ$ in $W$. Given all this, the uncertain property $(2)$ is
for any subset $Y$ of $Z$, we have that "$TY$ spans $TZ$" implies "$Y$ spans $Z$".
I tried to prove this as follows. Assume $TY$ spans $TZ$. Then $TY \,\cup\,(W - TZ)$ spans $W$. Each of the following spans $W$ also:
$$TY \,\cup\,(TV - TZ),$$
$$TY \,\cup\,(T(V - Z)),$$
$$T(Y \,\cup\, (V - Z)).$$
Then applying $(1)$, we find that $Y \,\cup\, (V - Z)$ spans $V$.
But why should this mean that $Y$ spans $Z$? I do not see that it does.
Property (1) together with $T$ surjective implies that $T$ is an isomorphism.
Indeed, for every subset $S$ of $V$ it is true that $TS$ linearly independent implies $S$ linearly independent (no condition of surjectivity of $T$ is necessary here).1
So take a basis of $W$; since $T$ is surjective, this basis has the form $TS$ for some subset $S$ of $V$. Now, by property (1), we know that $S$ spans $V$. Since it is linearly independent, it is a basis.
Since $T$ maps a basis to a basis, it is an isomorphism.
Footnotes
1 For whom is not used to “lists of elements” rather than “sets of elements”, the statement should be interpreted as: assume $S'$ is a set of elements in the range of $T$ $W$ and choose, for every $w\in S'$ a vector $v\in V$ with $Tv=w$; form the set $S$ of such elements. Then $S'$ linearly independent implies $S$ linearly independent.