# injective and surjective

How to prove that a composite function $f\circ g$ is bijective$?$

because i have two questions. if $f$ is injective and $g$ is surjective, can $g\circ f$ be both injective and surjective? because the question assumes both sets to be the same criteria ( $f$ injective, $g$ surjective) , so one question is whether $g\circ f$ is injective and the other is if $g\circ f$ is surjective.

i proved that if $f$ is injective then, $x=y$, $f(x)=f(y)$, then $g(f(x)) = g (f(y))$ so $g\circ f$ is injective. but is it true that if $f$ is injective and $g$ is surjective, then $g\circ f$ can also be surjective. its making me confused.

• For surjective, you want an element in the target set (where $g$ maps TO) to come from an element in the domain set (where $f$ maps FROM) in the case of $g\circ f$. So you notice that there must have been some $f(x)$ such that $g(f(x))$ is the element you wanted, so $x$ was the original element you needed. For injective, let $g(f(x))=g(f(y))$ and see where that takes you. You might find that a surjective $g$ is too weak. Or you might find it works. – Terra Hyde Sep 5 '15 at 5:39
• Your Injectivity proof is not correct – user118494 Sep 5 '15 at 5:41
• how is it not correct – Joseph Lopez Sep 5 '15 at 5:48
• See , to prove injectivity of $g\circ f$ , you have to show that , ; if $(g\circ f) (x)=(g\circ f)(y)$ then $x=y$. But what you have shown is that $x=y$ implies $(g\circ f) (x)=(g\circ f)(y)$ ,that is well-definedness,not injectivity – user118494 Sep 5 '15 at 5:53
• Keep one thing in your mind when you try to prove this kind of statement: $f \circ g$ is just another function. You start with let $h = f \circ g$, then prove $h$ is bijective using whatever conditions $f$ and $g$ have. – scaaahu Sep 5 '15 at 6:20

"if $f$ injective and $g$ surjective can $g\circ f$ be both injective and surjective?"

Sure, e.g. let $f:\{0\}\rightarrow\mathbb R$ and $g:\mathbb R\rightarrow\{0\}$.

But not always, e.g. let $g:\{0,1,2\}\rightarrow\{0,1\}$ be prescribed by $0\mapsto0$, $1\mapsto1$, $2\mapsto1$ and let $f:\{0,1\}\rightarrow\{0,1,2\}$ be prescribed by $0\mapsto1$, $1\mapsto2$.

Then $f$ is injective and $g$ is surjective.

However $g\circ f:\{0,1\}\rightarrow\{0,1\}$ is constant (and prescribed by $0\mapsto1$ and $1\mapsto1$).

It is evidently not injective and is not surjective.

• do u mean that it can be known as injective and also known to be surjective? so my statement that when f is injective and g is surjective, gof is injective is true? and the other statement when f is injective and g is surjective, gof is surjective is true? – Joseph Lopez Sep 6 '15 at 16:41
• My first example gives you an injective $f$, a surjective $g$ and a $g\circ f$ that is injective and surjective. So the answer on the question between quotes is: yes. My second example gives you again an injective $f$ and a surjective $g$, but now with a $g\circ f$ that is not injective and is not surjective. This together shows that injectivity of $f$ and surjectivity of $g$ does not exclude injectivity and/or surjectivity of $g\circ f$, but also does not guarantee injectivity and/or surjectivity of $g\circ f$. That's all I am saying. – drhab Sep 6 '15 at 16:49

drhab's answer is excellent. However, I want to add two more things that are true.

Theorem

Let $f: A \rightarrow B, g: B \rightarrow C$ be surjective functions. Then $g \circ f: A \rightarrow C$ is a surjection.

Proof:

Let $c \in C$. Then, there exists $b \in B$ such that $g(b) = c$ (because $g$ is surjective). Because $b \in B$, there exists $a \in A$ such that $f(a) = b$

Therefore, $c = g(f(a)) = g \circ f(a)$, leading us to conclude that $g \circ f$ is a surjection. $\quad \triangle$

Theorem

Let $f: A \rightarrow B, g: B \rightarrow C$ be injective functions.Then $g \circ f: A \rightarrow C$ is an injection.

Proof

Let $a,a' \in A$ such that $g \circ f(a) = g \circ f(a')$

equivalently, we can say:

$g(f(a)) = g(f(a'))$

and by injectivity of $g$:

$f(a) = f(a')$

and by injectivity of $f$

$a = a'$

Hence, $g \circ f$ is injective.$\quad \triangle$

From the two previous theorems, it immediately follows that the composition of $2$ bijective functions is bijective.