# 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 $S<T$.
Let $g:S \to U$ which is onto. Then $f_2 \circ g$ is onto $T$, a contradiction.
• Nice answer. I did not think that it is so easy. I totally forgot to consider its composition with $f_2$ function. Thanks a lot!
• $f_2 \circ g$, not $g \circ f_2$. Nov 9, 2017 at 2:35