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.

  • 1
    $\begingroup$ 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. $\endgroup$ – Terra Hyde Sep 5 '15 at 5:39
  • $\begingroup$ Your Injectivity proof is not correct $\endgroup$ – user118494 Sep 5 '15 at 5:41
  • $\begingroup$ how is it not correct $\endgroup$ – Joseph Lopez Sep 5 '15 at 5:48
  • $\begingroup$ 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 $\endgroup$ – user118494 Sep 5 '15 at 5:53
  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ 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? $\endgroup$ – Joseph Lopez Sep 6 '15 at 16:41
  • $\begingroup$ 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. $\endgroup$ – drhab Sep 6 '15 at 16:49

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


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


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. $\quad \triangle$


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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.