Proof that if g o f is Surjective(Onto) then g is Surjective(Onto)

[@chirayushah1608]

OMG I HAVE BEEN STRUGGLING WITH THIS FOR THE PAST WEEK. YOU MADE IT SOOOOOOOO EASY TO UNDERSTAND THANK YOU!!!!!!!!!

[@-AnweshaDas-]

YOU ARE INCREDIBLE YOU MADE IT SO EASY THANK YOUUUUU

[@TheMathSorcerer]

no problem!

[@Abs272b]

I hate pure math man it is not easy. Respect to all mathematicians

[@thebestisyettocome9317]

You're literally saving my degree

[@TheMathSorcerer]

❤️

[@robbertcox6170]

Thanks math sorcerer, big fan of your work🙌

[@null.dev.]

Been a good while since I did this at uni.. and after.. a goood while. At first I was like WAIT, just setting b = f(a) has to be proven... then watched again thinking and aha. The video seemed a bit messy to then be fairly concise :) cheers for the refresh. Additionally thankful for Bob Bobson's comment/example from 6 years ago :D.

[@lisstalikm8297]

It's still true* upside down? If f is onto, then f ◦ g is onto

[@dhruvmanohar7675]

bruh i love you saw this in between my test

[@TheMathSorcerer]

👍

[@TheMathSorcerer]

[@taeshunho6530]

What about if g is injective and we need to prove whether or not f is surjective?

[@chloepozderac712]

this is very informative however my problem kind of the reverse could you PLEASE help with this one? (iA and iB are the identity function on A & B respectively) Let f A: -> B and g B: -> A be functions such that f o g = iA and g is surjective. Prove that g o f = iB .

[@studysike]

Prove that f is onto iff h o f = k o f implies h = k. how to prove this and if A and B are finite sets with same number of elements then f:A->B is bijective if f is one one and onto?

[@ongkojoyo858]

Awww! Thank you so much sir, you made it so easy🙌🏻

[@BaraNoMatsuri]

Is this true for all A, B, C? Then how about f? Is it surjective for all A, B, C if gof is surjective?

[@bobbobson2061]

+BaraNoMatsuri f is not necessarily surjective. Example: A = {a, b} B = {m, n} C = {k} f: A → B with f(a) = f(b) = m g: B → C with g(m) = g(n) = k g∘f is surjective, f is not.

[@jk38589]

so does this mean that f needs to be surjective also?

[@MrAndersonAFK]

no. f can be non surjective, or surjective.

[@ruisamueltreves2484]

@@lucasigne8752 It doesnt matter if f is surjective or not because "a" is in the domain of the function f. Everything in the domain of a function has a corresponding thing in the range of that function; that's one defining thing of being a function. Thus, since f:A->B and "a" is in the set A, by assumption, there must exist an f(a)= "b" in the set B.

[@DantheAgario]

@@ruisamueltreves2484 No, because B is not the range of f. Is the codomain of f. So range of f can be smaller or equal to B

[@kuroshkabir136]

very good. thanks

[@parkermilligan7503]

I will hire you to be my tutor.

[@krishnageetha75]

But f(B)=y isn't an enough condition to be onto right every element is set c should have a preimage in B then I guess we have to take an inverse like some function h(y)=B I am confused pls help

[@johnny3475]

The y is any element in the codomain. All we have to show that there is an element in the domain, so b, such that g(b)=y. That's all you do.

[@Itsimane_]

What about f is it surjective or injective ??

[@alexandrashvydun8726]

it's not necessarily onto but idk if it's one-to-one or not

[@AU-yw2kq]

Ia the converse also true? If g(x) if surjective then gof(x) is also surjective?

[@mimiyep4235]

Yes

[@ibrahimolima1184]

I LOVE YOUUUUU

[@slimanemesbah8500]

thanks for the video~ what about f ( is it surjective )

[@nikhilkumar-rd3nz]

Need not be a surjective

[@وحیدسیروسی-ج8ز]

very nice

[@calvinlau9549]

What if f o g is surjective? Would f be surjective?

[@TheMathSorcerer]

The Math Sorcerer and the proof is the same, just rename f and g.

[@calvinlau9549]

The Math Sorcerer Thanks!

[@oleksandr8371]

this is only a one way implication as i am assuming? What would would be a counterexample to the following statement; if g is surjective, then g o f is surjective please provide an example when this is not true.

[@gnydnnk8384]

Definitely. Let A = {1,2}, B = {1,2,3,4}, C = {1,2,3,4}. Define f:A->B, g:B->C and so g o f: A->C Define f(x) = 2x, g(x) = x So g is surjective. (g o f)(x) = 2x which is not surjective since the image of (g o f) =/= C.

[@ermiles6472]

g composed with f

[@lisstalikm8297]

It's still true* upside down? If f is onto, then f ◦ g is onto

[@Abkibaar200paar]

Same doubt, did you find the answer??

[@DantheAgario]

When you say that f(a) = b you are assuming that is surjective. In fact, is the definition of surjectivity for f

[@TheMathSorcerer]

No I'm just calling it b

[@krishnageetha75]

@Eduardo that is the basic definition of a function , when you are defining f:A-B as a function then every element should have an image in B be it distinct or not I hope it's clear now

[@DantheAgario]

@@krishnageetha75 yes, thanks krishna:)

[@eggymari]

how did i get here

[@TheMathSorcerer]

Lol

[@eggymari]

@@TheMathSorcerer I think it was because I was revising ontological arguments proving God's existence today for my religious studies class at A-Level (UK qualification.)

[@goplay5936]

Can g be not surjective as well?

[@TheMathSorcerer]

If g o f is surjective then g is always surjective, that's what we proved.

[@jamespottex5197]

😅😅😅😅

[@anuragharsh2012]

@@TheMathSorcerer Can f be surjective as well if g:B->A

[@yesstar9963]

When there is only one video on a topic and you cant understand that video too😞😞😞

[@TheMathSorcerer]

LOL awww sorry man

[@TheMathSorcerer]

just go through the definition in each step, that's all we are using, just the definitions:)