Thread: Show that a function is injective

Suppose that f:A B and g:B C are mappings. Show that:

1. f and g are both injective g f is injective

I did it like this:

Let , and .

If f and g are injective x,y A, f(x)=f(y) x=y.

Therefore: f(a)=b and g(b)=c where a is unique to b and b is unique to c (one to one function by the injective property).



Let x=a:



Since a is unique to be and b is unique to c then is also a one to one function. Hence is injective.

__________________________________________________ _________

My issue here is the part where I mention that f(a)=b and g(b)=c. Is this a valid claim?

Originally Posted by Showcase_22

Suppose that f:A B and g:BC are mappings. Show that: 1. f and g are both subjective g f is injective

Look at the question you posted.
Should both be injective?
It seems that you are assuming that.

I don't know what "subjective" means here. If it was supposed to be "surjective", the statement is not true:
Let A= B= {x, y, z}, B= {a, b}, C= {p} f:A->B defined by f(x)= a, f(y)= a, f(z)= b, g:B->C defined by g(a)= p, g(b)= p. Then gof(x)= g(a)= p, gof(y)= f(a)= p, gof(z)= g(b)= p so both f and g are surjective but gof is not injective.

Since you asserted in your proof that f and g are both injective, is it possible that the problem supposed to be "if f and g are both injective prove that gof is injective"?

I did mean "surjective" instead of "subjective". However, I did mean what you posted back. I've changed the question now.

I was doing the second question (it's the same as this one with "surjective" instead of "injective") and I had "surjective" in my head while I was typing.

Thinking about it using diagrams, if g and f are surjective then gof cannot be injective as well.

So is my proof about right?

Hello,

it's not very difficult for the first one

use this property of injective functions :
f is injective <=> [f(x)=f(y) => x=y]

so start from where x and y belong to C.

so

but since g is injective, we have

and since f is injective, we have ....

Since f is injective we have x=y.

Therefore so gof is injective.

I think I get it now.