Prove that Im(TS) = Im(T) if $S$ is surjective and $T$ is any linear transformation Given that $S: U \to V$ and $T: V \to W$ prove the above...

Not sure where to start with this. I know that since $S$ is surjective, that $Im(S) = V$, but image isn't distributive right? I.e. $\mbox{Im}(TS) = \mbox{Im}(T).\mbox{Im}(S)$.

Suggestions appreciated.
Let $TS(x) \in \mbox{Im}(TS)$. Then, $TS(x) = T(S(x))$, so that $TS(x) \in \mbox{Im}(T)$ also.

The other way,let $T(y) \in \mbox{Im}(T)$. For this $y$, there exists $z$ such that $S(z) = y$ by surjectivity of $S$. Now, $TS(z) = T(y) \in \mbox{Im}(TS)$.

Hence, if $S$ is surjective, $\mbox{Im } T = \mbox{Im } TS$. Of course, if $T$ were surjective but $S$ is not, then this is obviously not true, for example $S$ could be the operator sending all elements to zero.
Let Im=f . f(T)= {y such that y=f(T(x)) for some x}={y such that y=f(T(S(x)) for some x}(because S is surjective)=f(TS).