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The definition of bijection I have learned is that there is bijection between 2 sets $A$ and $B$ iff there is a bijective function mapping from one of $A$ or $B$ to the other. I am wondering whether to show bijection between 2 sets, the injection and surjection has to come from the same function- if there is a surjective function $g: A \rightarrow B$ and injective function $f: A \rightarrow B$ where $f$ and $g$ are different functions, is this sufficient to show bijection between these sets? I believe it is, because the existence of $g$ shows that $|A| \le |B|$ and the latter shows that $|B| \le |A|$; however I am unsure due to the fact that the definition I have learned talked about a single function being both injective and surjective, rather than two functions between the sets possessing these qualities independently?

Princess Mia
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    I assume you're adopting the axiom of choice. You've got your inequalities flipped. But see the CBS. – blargoner Oct 14 '23 at 22:59
  • @blargoner are we able to invoke that theorem because the surjection from $A$ to $B$ shows injection from $B$ to $A$? – Princess Mia Oct 14 '23 at 23:12
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    Given the surjection $g: A \to B$, you can build an injection $h: B \to A$ by, for each $b \in B$, choosing $h(b)$ to be any element which is mapped to $b$ by $g$ (there is at least one, by surjectivity of $g$). (This uses the axiom of choice!) See also here. – Izaak van Dongen Oct 14 '23 at 23:49
  • This is the Schroder Bernstein theorem. The proof is not that easy. – Porky Oct 15 '23 at 01:35

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