Is my answer the same as the solution? [Linear Maps]

Question

I am trying this question on Linear Algebra Done Right:

In the solution, it said "Let $(w_1,...,w_m)$ be a basis of $W$" and do the work following.

My trial was

Suppose $T$ is surjective, prove $TS$ is an identity map on $W$. Let $w \in W$ and $v \in V$

$TS(w) = T(v) = w$

Then I said $TS$ is an identity map. I feel that it may be wrong. But I am not sure if this is ok.

Answer 1

Your answer is not the same as the solution because your answer doesn't solve the problem at all. You take a $w\in W$ and a $v\in V$ and you claim that $TS(w)=T(v)=w$ without explaining why would any of these equalities holds. Hint: in general, they don't.

Answer 2

In your solution you never say what $S$ is. So your solution isn't wrong in the sense that you haven't said anything that's wrong. But it's not right either because you haven't shown that $S$ exists (which is the whole point of the problem).

The issue for this problem is that it's tempting to define $S(w) = v$ if $T(v) = w$. However, this is only well defined if there is only one $v$ such that $T(v) = w$. If $T(v) = w$ and $T(v') = w$, you need $S$ to decide between $S(w) = v$ or $S(w) = v'$ and you need to be able to make this choice for all $w \in W$.