Prove that
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injective function $f:X\to Y$ exists
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surjective function $g:Y\to X$ exists
is equivalent.$(x,y\neq \emptyset)$
My approach
First,prove left to right.
If injective function $f:X\to Y$ exists, a set A can be defined like this.
$$A=\{f(x):x\in X\}$$
then it is sure that $A\subset Y$, and $|A|=|X|$
And, if we define function $g$ like
$$g(y)=\begin{cases}x & \text{if $y\in A$ and } f(x)=y,\\x_0& \text{if }y\notin A.\end{cases}$$
($x_0$ is a fixed element of $X$)
Then, $g$ is a surjective function, so surjective function $g:Y\to X$ exists.
Now,prove right to left.
If surjective function $g:Y\to X$ exists, $\forall x \in X$, we can select a element $y\in Y$ that $g(y)=x$, which is clear that we can pick a different $y\in Y$ for $\forall x \in X$
if we define function $f$ as $f(x)=\text{one element in Y that g(that element)=$x$}$, f is a injective function, so injective function $f:X\to Y$ exists.
But, I am not sure that my approach is right, especially on proving right to left.
Please check whether my approach is right, ways to improve it, and some other ways of proving this question.
Best Answer
It is correct but you have to quote Axiom of Choice for right-to-left implication.