Proof that if g o f is Injective(one-to-one) then f is Injective(one-to-one)

Proof that if g o f is Injective(one-to-one) then f is Injective(one-to-one). Given two functions f : A to B and g: B to C, we prove that if the composition g o f: A to C is an injective function then f is also an injective function.

[anonymousvevo]

i'm moroccan, i live in france, in the two f*** countries, i haven't found videos as good as yours, thank you so much

[taylormclean]

There is a nuance that tripped me up initially, that I'd like to point out. We can go from f(x)=f(y) => g(f(x))=g(f(y)) because we assume g is a function, therefore cannot map one value to many.

[shanegalvin]

Suppose g of f is injective and (for contradiction) assume that f is not injective. Since f is not injective, there exists a, b in A such that f(a)=f(b), where a=/=b. 

Then, g(f(a))=g(f(b)), where a =/= b. 

i.e. we've show that there exists two elements in A that map to the same element in C. This is a contradiction. Therefor, if g of f is injective, so too is f.

[kumarshanti]

When you search for a topic and you get this channel videos on it, sorcerer does exist to solve your problem. Love your contents

[jesselam]

I'm not sure if I get this proof, from my understanding all you have proven is that the function f can either be a surjective or injective function but it doesn't strictly prove one or the other. 
Since f(x) = f(y) can still hold true if f was surjective, so I'm not sure why you can just assume x = y.

[trungtranang]

Love it, how about injectivity of g? Can you proof g is injective or not injective in this case ? Thanks a lot

[angthinhtuongminh]

Proof only correct given that g(x) is injective ? f(a)=f(b) doesn't necessary mean that g(f(a))=g(f(b)) given that g(x) can be surjective

[zaylo]

Your explanation is the best! Mind blowing XD

[flebus]

I don't get this, what if x and y are different values and f(x) just happens to equal f(y)?

[imdadbhuiyannayem]

Consider functions f : A→B and g : B→C Prove the following: 1) if f and g are one-to one, then the composition function gof is one-to-one. ii) If f and g are onto functions, then gof is an onto function.
solve this pls

[doaaserageldin]

neat proof, thanks but can you do a graph? i dunno why i can't proof without graphing

[kinwaiwong]

you were trying to prove if p then q but you suppose p and suppose q (which actually is something you should show). It is not a legitimate proof.

[addajs]

But i think A should equal to B for that being true ??

[thomaskingston]

How does this prove that f is injective?

[luisrodriguez-mxme]

dont you beg the question by saying at the start that f(a)=f(b) because that's the thing you're trying to show?

[salmaziadi]

Consider gof is surjective and f is surjective. How can we prove that f is injective