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 .
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.
+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.
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?
@@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.
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
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.
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.
@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
@@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.)
OMG I HAVE BEEN STRUGGLING WITH THIS FOR THE PAST WEEK. YOU MADE IT SOOOOOOOO EASY TO UNDERSTAND THANK YOU!!!!!!!!!
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no problem!
I hate pure math man it is not easy. Respect to all mathematicians
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Thanks math sorcerer, big fan of your work🙌
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 .
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.
It's still true* upside down? If f is onto, then f ◦ g is onto
What about if g is injective and we need to prove whether or not f is surjective?
bruh i love you saw this in between my test
👍
Is this true for all A, B, C? Then how about f? Is it surjective for all A, B, C if gof is surjective?
+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.
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?
so does this mean that f needs to be surjective also?
no. f can be non surjective, or surjective.
@@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.
@@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
thanks for the video~ what about f ( is it surjective )
Need not be a surjective
very good. thanks
Awww! Thank you so much sir, you made it so easy🙌🏻
What about f is it surjective or injective ??
it's not necessarily onto but idk if it's one-to-one or not
Ia the converse also true?
If g(x) if surjective then gof(x) is also surjective?
Yes
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
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.
What if f o g is surjective? Would f be surjective?
The Math Sorcerer and the proof is the same, just rename f and g.
The Math Sorcerer Thanks!
very nice
I will hire you to be my tutor.
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.
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.
It's still true* upside down? If f is onto, then f ◦ g is onto
When you say that f(a) = b you are assuming that is surjective. In fact, is the definition of surjectivity for f
No I'm just calling it b
@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
@@krishnageetha75 yes, thanks krishna:)
g composed with f
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how did i get here
Lol
@@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.)
When there is only one video on a topic and you cant understand that video too😞😞😞
LOL awww sorry man
just go through the definition in each step, that's all we are using, just the definitions:)
Can g be not surjective as well?
If g o f is surjective then g is always surjective, that's what we proved.
😅😅😅😅
@@TheMathSorcerer Can f be surjective as well if g:B->A