Problem 15 from Herstein's book Given two sets $S$ and $T$ we declare $S<T$ if there is a mapping of $T$ onto $S$ but no mapping of $S$ onto $T$. Prove that if $S<T$ and $T<U$ then $S<U$.
My Proof: Since $S<T$ then $\exists$ onto mapping $f_1:T\to S$ and since $T<U$ then $\exists$ onto mapping $f_2:U\to T$ then mapping $f_1\circ f_2:U\to S$ is also onto mapping. We have done with the first part of proposition.
But how to prove that no mapping from $S$ to $U$ is onto. I have tried to do it by contradiction but no results.
Can anyone help please with that.

Suppose that there exist an onto map $f:S\to U$. Since $T<U$, there exist an onto map $f_0:U\to T.$ So $f_0\circ f:S\to T$ is onto and, what is a contradiction since $S<T$.
