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I know that in set theory it should be the case that if $\vert X\vert\geq \vert Y\vert$ and $\vert Z\vert\geq \vert W\vert$, then $$ \vert X^Z\vert\geq \vert Y^W\vert .$$ For this I want to map a function $f:W\to Y$ injectively to a function $g:Z\to X$. It seems natural that I construct the map $g$ from the conditions on $\vert X\vert\geq \vert Y\vert$ and $\vert Z\vert \geq \vert W\vert$, however I can not use the fact that $\vert Z\vert\geq \vert W\vert$ directly without the axiom of choice. My proof is as follows:

Since $\vert Z\vert \geq \vert W\vert$, there exists a surjective function $G:Z\to W$. Since $\vert X\vert \geq \vert Y\vert$, there exists an injective function $F:Y\to X$. Then the map $\Phi:Y^W\to X^Z$ is an injective function defined by $\Phi(f)=F\circ f \circ G$. However this proof requires that there exists such a surjective map, which I understood to not be necessarily the case if we don't assume the axiom of choice.

My question is what would be a correct injective mapping from $Y^W$ to $X^Z$, since I'm pretty sure that the property I want to show is true without assuming the axiom of choice.

Keen-ameteur
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  • Do you mean there is an injection $G : W \to Z$? – balddraz Nov 12 '20 at 12:00
  • @0XLR I think I corrected it accordingly now. I don't know how to a diagram in latex to draw my intention. – Keen-ameteur Nov 12 '20 at 12:11
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    Note that $\geq$ is NOT defined by "there is a surjection", it is defined by "there is an injection". You're doing it wrong, so you're appealing to choice, but the wrong part is entirely redundant. – Asaf Karagila Nov 12 '20 at 12:13
  • @AsafKaragila My original intention was that "there is an injection hence there is a surjection by Choice". But as I was reminded in the other thread, the existence of the surjection does not require the axiom of choice, which solves my problem. – Keen-ameteur Nov 12 '20 at 12:30

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