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I am not fluent in MathJax nor math lingo, so please bear with me. As I understand, for these declarations (consider these sets as ordered):

$L=\{a, b, c\}, l \in L$

$N=\{1, 2, 3\}, n \in N$

the following mapping is considered bijective, as it's both injective and surjective (what is enough, I uderstand):

$l_i \mapsto n_i \iff n_i \mapsto l_i$.

But how about this mapping?

$l_1 \mapsto n_2, l_2 \mapsto n_3, l_3 \mapsto n_1$ and $n_1 \mapsto l_1, n_2 \mapsto l_2, n_3 \mapsto l_3$.

It is injective, it is surjective (unless I am wrong, that is), but mapping from L to N and then from N to L won't give me the same element of L.

So: 1) is injective and surjective requirement sufficient to say, that mapping is bijective? 2) is the second mapping bijective? If not, is there a name for it (other than injective and surjective simultaneously). 3) If answer to first question is "no", then what is/are other requirement(s)?

Thank you for your insights! :-)

Cromax
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    Both mappings in the second example are bijective, but one is not the inverse of the other. – blamocur Jan 28 '23 at 21:51
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    Not all bijections are involutions. You should find all 6 bijections between these two sets and consider iterations of them. – CyclotomicField Jan 28 '23 at 21:55
  • Thank you for explaining, guys, so it was the definition I got wrong. Anyway receiving -1 for the first question just doesn't give much courage to ask for things, even if the answers provide useful clarifications. Guess it may be wrong SX for these kind of questions… – Cromax Jan 28 '23 at 22:43
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    «is injective and surjective requirement sufficient to say that mapping is bijective» you ask: being bijective means being simultaneously injective and surjective. Your question is not different than "is white white?". – Mariano Suárez-Álvarez Jan 28 '23 at 23:35
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    Beware that in each of your 2 examples you are not defining one bijection but two: one from $L$ to $N$ and another one from $N$ to $L.$ Of course one can identify such a pair of maps with a map from $L\cup N$ to itself but I have the (possibly wrong) feeling that it is not what you intended. – Anne Bauval Jan 28 '23 at 23:35
  • Down votes are this community's notion of a welcome, ignore them. – Mariano Suárez-Álvarez Jan 28 '23 at 23:35
  • @Cromax votes on a question will decide which answers are given first during a search. If a question already has a good answer elsewhere in the site some users will downvote repeat questions to draw less attention to them. Worry more about close votes as those indicate a serious problem with the question. In this case, you have zero which is excellent. Carry on. :) – CyclotomicField Jan 28 '23 at 23:39
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    @AnneBauval Now I get it — previously I had this wrong impression, that bijection somehow implies two-way relation/mapping. I was obviously wrong, hence I had doubts. It turned out, as CyclotomicField pointed, I was looking for involutory mapping—I am not mathematician and that's why my math vocabulary is limited. Thank you! :-) – Cromax Jan 29 '23 at 02:32
  • @MarianoSuárez-Álvarez I politely disagree. I'd rather say, that my question is congruent to "is white white" but distinctively different. It's not different than "is white being a sum of all frequencies in visible spectrum". The first is a tautology, the other a definition (well, a verifying question) of not ignotum per ignotum kind. And thank you for the clarification as to conditions required by function to be bijective. :-) – Cromax Jan 29 '23 at 02:40

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