A Fellow here at MSE recently wrote a post asking how to Prove if $f:X\rightarrow Y$ is surjective then $|X|\ge|Y|$.
I thought as a challenge it might be a good exercise to try to prove this claim in a different way the following is my attempt at constructing a proof is it correct?
Proof. We construct the proof by recourse to Strong-Induction on $|Y|$.
Let $n$ be an arbitrary natural number. Assume that the claim in question is true for an arbitrary natural number $k$ strictly less that $n$ that is given any sets $V$ and $W$ where $|W| = k<n$ and an arbitrary function $h:V\to W$ we have $|V|\geq |W|$ whenever $h$ is surjective.
We now prove the result for $n$ . Let $X$ and $Y$ be arbitrary sets such that $|Y| = n$ and let $f$ be a surjective map from $X$ to $Y$.
We start by partioning the set $Y$ so that $Y = Y'\cup\{\beta\}$, evidently $|Y'| = n-1$ . Observe that since $f$ is surjective so must be the functions $g:f^{-1}(Y')\to Y'$ and $h:f^{-1}(\{\beta\})\to\{\beta\}$ defined such that $g(x) = f(x)$ for all $x\in f^{-1}(Y')$ and $h(x) = f(x)$ for all $x\in f^{-1}(\{\beta\})$.
Then by considering these functions together with the inductive hyposthesis it follows that $|f^{-1}(Y')|\geq |Y'|$ and $|f^{-1}(\{\beta\})|\geq |\{\beta\}|$ furthermore since $Y'\cap\{\beta\} = \varnothing$ it follows that $f^{-1}(Y')\cap f^{-1}(\{\beta\}) = \varnothing$ consequently $$|X| = |f^{-1}(Y')\cup f^{-1}(\{\beta\})| = |f^{-1}(Y')|+|f^{-1}(\{\beta\})|\ge|Y'|+|\{\beta\}| = (n-1)+1 = n$$ completing the inductive step.
$\blacksquare$
LINK TO ORIGINAL POST: Prove if $f:X\rightarrow Y$ is surjective then $|X|\ge|Y|$
