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Start Learning Sets - Part 6 - Injectivity, Surjectivity and Bijectivity 

The Bright Side of Mathematics
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I hope that this helps students, pupils and others. Have fun!
(This explanation fits to lectures for students in their first and second year of study: Mathematics for physicists, Mathematics for the natural science, Mathematics for engineers and so on)

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21 июл 2024

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Комментарии : 27   
@MrOvipare
@MrOvipare 3 года назад
There is something beautifully simple about Set Theory. I'm sure it gets really hardcore in the axiomatic approach, but the core ideas are very reasonable and simple, fundamental.
@angelmendez-rivera351
@angelmendez-rivera351 3 года назад
Relating this to the concepts in the previous video, you can also give an alternative definition. A function f : A -> B is injective iff, for every singleton {y} in B, the preimage of {y} under f is either also a singleton {x} in A, or else the empty set {}. A function f : A -> B is surjective iff, for every singleton {y} in B, the preimage of {y} under f is a nonempty set in A. A function f : A -> B is bijective iff, for every singleton {y} in B, the preimage of {y} under f is always exactly a singleton {x} in A, i.e, iff f is injective and surjective. I think this is a more intuitive way of defining injectivity, surjectivity, and bijectivity, without losing any of the rigor, and the reason I think it is more intuitive is because the only concepts you need to understand is the concept of a preimage, the concept of a singleton, and the concept of the empty set, all of which are elementary notions and precede the concepts in these definutions. Another way to put it is that, if every singleton in the codomain of a function has preimage with cardinality C satisfying C =< 1 (C = 1 or C = 0), then the function is injective, while if every singleton in the codomain has preimage with cardinality C satisfying C >= 1, then it is surjective. If it is both injective and surjective, then C >= 1 and C =< 1, meaning C = 1, and so it is bijective. However, the advantage of the definitions I presented above is that are able of capturing this rather intuitive concept without the necessity of invoking the definition of cardinality, which is actually preceded by the definition of a bijection.
@ichkaodko7020
@ichkaodko7020 2 года назад
are u flexing or wut?
@spyrosmanolidis8516
@spyrosmanolidis8516 28 дней назад
It makes a lot of sense too because, by the definition we saw in a previous video, for a function f: A → B, f(x) = y and f(x) = ỹ means that y = ỹ, because you cannot have an x which maps to two different y. Likewise, the other way around, you have to have exactly one arrow, you cannot have more. I know that's probably common sense but I felt good coming onto that realization (boy will I be embarrassed if this is wrong), so yeah :)
@blackhitler8572
@blackhitler8572 Год назад
your voice is so soothing and trusting
@angelmendez-rivera351
@angelmendez-rivera351 3 года назад
Also, technically, bijectivity and invertibility are different concepts, and are defined differently. A function f is invertible, by definition, iff it has a right-inverse g, a left-inverse h, and g = h. However, the reason we identify bijectivity with invertibility, and vice versa, is thanks to a powerful theorem in set-theoretic functional analysis that says that f is bijective iff f is invertible.
@gurashishanand3670
@gurashishanand3670 9 месяцев назад
Thanks for the help
@bkpunt8633
@bkpunt8633 11 месяцев назад
thank you!!!
@brendawilliams8062
@brendawilliams8062 2 года назад
Thx.
@HelloWorlds__JTS
@HelloWorlds__JTS Год назад
Great video, as usual! Do you have another video where you clarify the case for maps where not every x∈A gets mapped to any y∈B? This seems like it would be relevant in, e.g., de Rham cohomology.
@brightsideofmaths
@brightsideofmaths Год назад
I don't have a video yet, but it's a good idea :)
@ahmedamr5265
@ahmedamr5265 9 месяцев назад
Thanks for the amazing explanations. One question: can we use the biconditional in the definition of the injective?
@brightsideofmaths
@brightsideofmaths 9 месяцев назад
Yes, absolutely! However, that is not needed because the other direction is already given in the definition of a map.
@celestialowl8865
@celestialowl8865 Год назад
Hey, I pretty comfortably understood everything here and definitely feel like im making good progress, but I did have one question. How do you establish a co-domain? The last example supposes the co-domain is just the set of squared values, but that feels like a very easy way to always convince yourself that a map is surjective. If a map is say, f(n) -> 2^n, how would I consider the co-domain? Or is the co-domain synonymous with the set of all outputs..in which case how is there a map that isn't surjective.
@brightsideofmaths
@brightsideofmaths Год назад
The codomain is given with the definition of the function. In other words: you choose is big enough such that the definition you want for the function works.
@ish5312
@ish5312 3 года назад
Is there a place where I can test my understanding on these ? Like maybe something like practice questions
@zealous919
@zealous919 2 года назад
Yes, but you have to pay a euro or something like that I think
@iliasaarab7922
@iliasaarab7922 2 года назад
@@zealous919 do you have a link?
@theblinkingbrownie4654
@theblinkingbrownie4654 Год назад
Quizzes are in the desc
@ThemJazzyBeats
@ThemJazzyBeats Год назад
The map from A to B, where A is the set of mathematical notations and B is their meaning, is not an injective map, given that for the same symbol (f^-1), we can have 2 meanings (the preimage and the inverse map) :p
@brightsideofmaths
@brightsideofmaths Год назад
It's not even a map with this definition ;)
@ThemJazzyBeats
@ThemJazzyBeats Год назад
@@brightsideofmaths Would you care to elaborate on why it's not even a map ? Is it because not all of A (so not all mathematical notations) has a correspondance in B ?
@brightsideofmaths
@brightsideofmaths Год назад
You already said it before: each x one the left gets multiple y values on the right. For a map each x gets one y. You can watch the former video about that.
@ThemJazzyBeats
@ThemJazzyBeats Год назад
@@brightsideofmaths Oh yeah woops I meant B to A then :p
@gasmaskgaming5865
@gasmaskgaming5865 3 года назад
Sir love u r videos but explain the concept as point of starter
@angelmendez-rivera351
@angelmendez-rivera351 3 года назад
He did explain it
@spyrosmanolidis8516
@spyrosmanolidis8516 28 дней назад
its a series, you have to watch the videos from the start. I think he has the playlist on his channel
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