# Thread: Show that a function is injective

1. ## Show that a function is injective

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

1. f and g are both injective $\Rightarrow$ g $\circ$f is injective
I did it like this:

Let $a \in A$, $b \in B$ and $c \in C$.

If f and g are injective $\forall$ x,y $\in$A, f(x)=f(y) $\Rightarrow$ 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).

$g \circ f(x)=g(f(x))$

Let x=a:

$g \circ f(a)=g(f(a))=g(b)=c$

Since a is unique to be and b is unique to c then $g\circ f(a)$ is also a one to one function. Hence $g \circ f$ is injective.

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

2. 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.

3. 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"?

4. 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?

5. 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 $g \circ f(x)=g \circ f(y)$ where x and y belong to C.

so $g(f(x))=g(f(y))$

but since g is injective, we have $f(x)=f(y)$

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

6. Since f is injective we have x=y.

Therefore $gof(x)=gof(y) \Rightarrow x=y$ so gof is injective.

I think I get it now.