# Thread: injection & surjection proofs

1. ## injection & surjection proofs

Here is my problem...
Let A, B, and C be nonempty sets and let f: AB and g: BC be functions.
a) Prove that if g f is an injection, then f is an injection.
b) If g f is an injection, then must g be an injection? Support your answer with a proof or counterexample.
c) Prove that if g f is a surjection, then g is a surjection.
d) If g f is a surjection, then must f be a surjection? Support your answer with a proof or a counterexpample.

Any help is greatly appreciated! Thanks!

2. Originally Posted by ElenaS519
Here is my problem...
Let A, B, and C be nonempty sets and let f: AB and g: BC be functions.
a) Prove that if g f is an injection, then f is an injection.
b) If g f is an injection, then must g be an injection? Support your answer with a proof or counterexample.
c) Prove that if g f is a surjection, then g is a surjection.
d) If g f is a surjection, then must f be a surjection? Support your answer with a proof or a counterexpample.

Any help is greatly appreciated! Thanks!

Let : f(x)=f(y) ====> g(f(x))=g(f(x)) ====.> gof(x)=gof(y) =====> x=y

since gof injective,hence f is injective

let zεC THEN there exists xεA SUCH that gof(x) = z since gof surjective.

BUT xεA THEN there exists yεB SUCH that f(x) =y ,hence gof(x)=g(f(x))=g(y)=z hence g surjective

3. Let:

f:{1,2}=A $\Longrightarrow$ {a,b,c}=B, and

g:{a,b.c}=B $\Longrightarrow$ {d.e}=C

AND such that:

f(1) = a
f(2) = b

g(a) = d
g(b) = e
g(c} = e

and now form:

gof(1) = g(f(1)) =g(a)= d
gof(2) = g(f(2)) =g(b) =e

From the above we see that:

gof is injection .surjection

f is injection but g is not ,because g(b)=g(c) and $b\neq c$

g is surjection but f is not ,because there is no element in A ,x such that f(x) =c