Proof: Composition of Surjective Functions is Surjective | Functions and Relations

preview_player
Показать описание
Let g and f be surjective (one to one) functions, where g maps A to B and f maps B to C. Then the composition fog, which maps A to C, is also surjective. We'll prove this result about surjective functions and their compositions in today's lesson!

The proof is very straightforward, and merely requires us to apply the definition of surjective functions a few times! Remember a function is surjective if every element in the codomain is mapped to by some element in the domain. And in this lesson we'll prove that function composition preserves surjectiveness!

The proof in this video, in combination with the proof in the above lesson, proves that the composition of bijective functions is also bijective.

I hope you find this video helpful, and be sure to ask any questions down in the comments!

********************************************************************
The outro music is by a favorite musician of mine named Vallow, who, upon my request, kindly gave me permission to use his music in my outros. I usually put my own music in the outros, but I love Vallow's music, and wanted to share it with those of you watching. Please check out all of his wonderful work.

********************************************************************

+WRATH OF MATH+


Follow Wrath of Math on...

Рекомендации по теме
Комментарии
Автор

I CAN'T THANK YOU ENOUGH FOR THIS VIDEO!!

krishnagajjar
Автор

can someone Please share the video that he mentioned about the converse of this proof, and thanks a lot really it helped me a lot in my studies

mubarakbintarsh
Автор

Thank you for the very helpful video♥️

anupaalwis
Автор

This was extremely helpful! You're such a good teacher! Could you do a video on showing the composition of bijective functions is bijective?

isabelfrance
Автор

hello.. is it onto or one to one which called as surjecttive?

Opena-fxqf
Автор

do you have one that proves that the Composition of injective Functions is injective

computergenius
Автор

Why is it f o g? I shouldve been g o f

shayorshayorshayor