The Composition of Surjective(Onto) Functions is Surjective Proof

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The Composition of Surjective(Onto) Functions is Surjective Proof.
I included some pictures in the proof with the hope that perhaps it makes more sense.

[@gainfulanalytics]

Thanks for this wonderful little example! Very helpful!!!!!

[@TheMathSorcerer]

I'm glad it helped:)

[@jonathanjayes]

You're truly a wizard, thank you!

[@TheMathSorcerer]

glad it helped!

[@livissocool13]

Thanks so much!!!! This set (haha get it) of videos has saved me this unit.

[@sebastian1ballack]

Thank you... you are the best.

[@TheMathSorcerer]

np!!

[@prateek0536]

Very helpful thank you sir.Loved it

[@TheMathSorcerer]

awesome:)

[@bashair7614]

Thank u so much u r amazing 👍🏼

[@livepoetry6230]

Thank you Sir

[@FPrimeHD1618]

Really nice video. Having the map drawn out really helps. I am finding the hardest part for me in this class isn't the logic, it's turning my logic into words. How was you experience when you first started writing math proofs?

[@TheMathSorcerer]

It's tough for everyone!

[@FPrimeHD1618]

The Math Sorcerer Then there is that one kid in class who is getting a perfect on everything, and you can't help but think, "they will probably have a manifold named after them one day." lol.

[@amyagold4334]

this has confused me all semester thanks a lot

[@TheMathSorcerer]

+Amy A Gold np, I am glad it helped

[@TheAcwebs]

Amazing!!!!

[@TheMathSorcerer]

glad it helped!

[@elgabos99]

Shouldn´t the compositon of g: A to B and f: B to C be "g o f" instead of "f o g" ?

[@TheMathSorcerer]

[@hasek747]

I have a question, hope you can help. It says that "Since g is onto, there exists a little a belonging to big A so that g(a)=b" Isn't this true for all functions, though - not just surjective functions? Doesn't any function with a non-empty domain fulfill the requirement that there exists a number belonging to its domain such that this number maps to a number in the codomain? Thank you

[@TheMathSorcerer]

that is not the definition of onto, and you are correct in saying "any function with a non-empty domain fulfill the requirement that there exists a number belonging to its domain such that this number maps to a number in the codomain". Anyways what you stated initially is not the definition of onto. Hope this helps.

[@TheMathSorcerer]

f:A->B is onto if for every b in B we can find a in A such that f(a) = b. Not every function satisfies this property. Take f:[0, inf) -> R, f(x) = sqrt(x), -1 is in R but there is no x such that f(x) = -1, if it was we would have sqrt(x) = -1, which doesn't make sense. Hope this helps.

[@hasek747]

Wow, thank you for such a prompt reply! Sorry, I didn't mean to suggest that this was the definition of onto. What I am having a difficulty understanding is why we are saying that: "Since g is onto, there exists a little a belonging to big A such that g(a)=b". In other words, I do not understand this implication that "because g is onto" then "there exists a little a belonging to big A such that g(a)=b". It sounds - pardon the word - trivial, in the same way that saying "because g is an exponential function, there exists a little a belonging to big A such that g(a)=b". Sorry if I'm complicating things here, just trying to understand.

[@hasek747]

In other words, why don't we just say: "Since g is onto, for every little c belonging to big C, there exists a little b belonging to big B such that g(b) = c". Shouldn't this be what we're going for? and if not, then why? Thank you! Then we could say the same further: "Since f is onto, for every little b belonging to big B, there exists a little a belonging to big A such that f(a)=b" Then: (g o f)(a) = g(f(a)) = g(b) = c, therefore (g o f)(a) is surjective.

[@TheMathSorcerer]

it's because b is in B, so there is a in A such that g(a) = b

[@ShivamSingh-jn8ju]

oh thanx man!

[@TheMathSorcerer]

np at all!

[@mariodelrio6865]

Shouldn’t it be E! for that x€X at the recall? Like there’s just one element x from X that goes to an specific y, right?

[@TheMathSorcerer]

not necessarily no, it doesn't have to be unique

[@mariodelrio6865]

@@TheMathSorcerer okay i see, thanks, you’re awesome man! Your videos are so helpful!

[@TheMathSorcerer]

👍

[@donttalktome313]

Thanks for this. Our exam is today

[@TheMathSorcerer]

Np good luck😄

[@krumpy8259]

let f: X → Y, T ⊆ Y can you prove that these following 3 statements are equivalent: i) f is surjective , ii) f(f^-1(T))=T for all T ⊆ Y , iii) f(f^-1(Y))=Y . Thanks!

[@TheMathSorcerer]

Interesting! I actually might do this soon, looks cool, thank you very much for the suggestion:)

[@MichaelBrashier]

what is F is onto but G is not? is F o G onto?

[@TheMathSorcerer]

+Michael Brashier not necessarily!

[@MichaelBrashier]

thanks. That validates an answer to a problem i was working on.

[@ebomegaedits4701]

Is it the same for f*g? Thanks

[@TheMathSorcerer]

similar proof yes

[@TheMathSorcerer]

wait do you mean "times"?

[@ebomegaedits4701]

Is it the same for g*f

[@TheMathSorcerer]

similar proof yup

[@TheMathSorcerer]

wait do you mean "times"?

[@intuitiveclass6401]

Fucking amazing