Prove if $f:X\rightarrow Y$ is surjective then $|X|\ge|Y|$
My work:
Let $X,Y$ sets.
Suppose $|X|\leq |Y|$ then exists a function $g:X\rightarrow Y$ injective.
Here I'm stuck, can someone help me?
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1The conclusion of the implication should be $|X|\ge |Y|$ not $|X|>|Y|$ – atifcppprogrammer Apr 15 '18 at 22:49
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true, thanks @AtifFarooq – rcoder Apr 15 '18 at 22:50
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1See also: There exists an injection from $X$ to $Y$ if and only if there exists a surjection from $Y$ to $X$. (and other posts linked there). – Martin Sleziak Apr 16 '18 at 03:02
2 Answers
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The statement is false. Given any set $X$, the identity from $X$ onto itself is surjective. However, it's not true that $|X|>|X|$.
José Carlos Santos
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As remarked in the previous answer, you want $\geq$.
for each $y \in Y$, consider$f^{-1}(\{y\})$ and choose a unique $x$ in there. Taking the uniion of all these, there is a collection $U \subset X$ so that $f(U)=Y$, and is a biijection, so there is a bijection $g:Y \to U \subset X$
Andres Mejia
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