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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.

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2 Answers 2

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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 $S<T$.

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Let $g:S \to U$ which is onto. Then $f_2 \circ g$ is onto $T$, a contradiction.

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    $\begingroup$ I answered 22 seconds before you did. :-) $\endgroup$
    – Filburt
    Commented Nov 8, 2017 at 17:34
  • $\begingroup$ 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! $\endgroup$
    – RFZ
    Commented Nov 8, 2017 at 17:36
  • $\begingroup$ @Filburt you got my +1 $\endgroup$
    – gt6989b
    Commented Nov 8, 2017 at 17:38
  • $\begingroup$ @gt6989b So you got mine $\endgroup$
    – Filburt
    Commented Nov 8, 2017 at 17:39
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    $\begingroup$ $f_2 \circ g$, not $g \circ f_2$. $\endgroup$ Commented Nov 9, 2017 at 2:35

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