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Suppose $f : A \to B$ is one-to-one, and there is another function $g : B \to A$ which is also one-to-one. We don’t assume anything in particular about the relationship between $f$ and $g$. Are $f$ and $g$ necessarily bijections?

malte
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    I don't think so, no; maybe as long as they're finite/countable. But you can always use them to construct a bijection. See the Schröder-Bernstein theorem https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem and its proof – SV-97 Oct 08 '21 at 12:20
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    If you want an example... let $A=B=\Bbb N$ and for each $n\in \Bbb N$ let $f(n)=g(n)=2n$. These are clearly both injections (as if $2a=2b$ then necessarily $a=b$) yet neither are a surjection (there is no natural number $n$ such that $2n = 3$ for instance). – JMoravitz Oct 08 '21 at 12:23
  • Another example is the inclusion of $\mathbb{R}^+$ into $\mathbb{R}$. Clearly this map is injective but not surjective - yet both sets are bijective. – SV-97 Oct 08 '21 at 12:27

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