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I have a small question about this problem, the problem says:

Given 2 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 $S<U$

So the question when I say mapping of $T$ onto $S$ but no mapping of $S$ onto $T$, are two functions or only one ?

Altaid
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    There exists one function from $T$ to $S$ that is surjective; also, there does not exist a function from $S$ to $T$ that is surjective. – Greg Martin Dec 03 '21 at 02:26
  • It means there is a function $f:T\to S$ which is onto, but no function $g:S\to T$ which is onto. – Thomas Andrews Dec 03 '21 at 02:34
  • ty guys, I can prove the statement now. – Altaid Dec 03 '21 at 02:39
  • Just to be clear, you have to prove two things: (1) there is a function from $U \to S$ which is surjective, and (2) there cannot be a function from $S \to U$ which is surjective. Make sure you have shown both. – Paul Sinclair Dec 03 '21 at 14:44
  • Yeah, for part (1) I use the hypothesis and the theorem of the composition of 2 subjective functions and for (2) I see if $g$ no exist is the same as for all functions from $S$ to $T$ this function is not subjective in this case, in particular, Let two functions $f:S\to T$ and $g:T\to U$ such that their composition is not subjective we follow the result. – Altaid Dec 03 '21 at 20:20

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