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If $\sigma$: $A$ $\rightarrow$ $B$ was a mapping which was one to one, and had an inverse $\sigma$$^{-1}$: $B$ $\rightarrow$ $A$ which is also one to one, then are they both bijective mappings? I'm almost certain that this is true, and is pretty much just a different way of saying the mapping is bijective, but I'm a bit ill at the moment and so I don't trust my brain too much on this.

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  • $\begingroup$ In fact, assuming by inverse you mean a two sided inverse (both $\sigma\sigma^{-1}$ and $\sigma^{-1}\sigma$ being the appropriate identity) the existence of such an inverse implies that $\sigma$ is one-to-one and onto. $\endgroup$ – JHance Aug 20 '14 at 16:56
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If $\sigma^{-1}$ is defined on all of $B$, this means that $\sigma$ is onto too. Hence, $\sigma$ is a bijection. Inverses of bijections are also bijections.

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  • $\begingroup$ Yes, I thought so, I just for some reason had a nagging suspicion that that wasn't the same as sigma being onto...I know that the question was simple, but I couldn't let that suspicion go $\endgroup$ – Nethesis Aug 20 '14 at 16:51
  • $\begingroup$ If $\sigma $ wasn't onto, we would only have $\sigma^{-1}: \mathrm{Im}(\sigma) \to A$. $\endgroup$ – Ivo Terek Aug 20 '14 at 16:57
  • $\begingroup$ I know, that was the exact same logic I was using, it's just that right now with my head pounding so much I could say 2+2=4 and still not believe it $\endgroup$ – Nethesis Aug 20 '14 at 16:59
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Let $\sigma \subseteq A \times B$. Suppose $\sigma$ is 1-1 in the sense: for every $a$ there is at most one $b$ with $(a,b) \in \sigma$; and suppose $\sigma^{-1}$ is 1-1 in the sense: for every $b$ there is at most one $a$ with $(a,b) \in \sigma$. Then $\sigma$ is a bijection from the domain of $\sigma$ to the range of $\sigma$.

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