If g∘f Is Bijective, Is f Injective and g Surjective?

Suppose $f:A\to B$ and $g:B\to C$, so the composition $g\circ f:A\to C$ is defined. If $g\circ f$ is bijective, then $f$ is injective and $g$ is surjective. The two conclusions follow directly from the injectivity and surjectivity parts of bijectivity.

The function types matter: the codomain of $f$ must fit the domain of $g$, and the composite is considered as a map into $C$. In particular, saying the composite is bijective means it is onto all of $C$, not merely onto its image.

Why f is injective

A function is injective if equal outputs can only come from equal inputs. Take any $x_1,x_2\in A$ and suppose

$$f(x_1)=f(x_2).$$

Applying $g$ to both sides gives

$$g(f(x_1))=g(f(x_2)),$$

or equivalently,

$$(g\circ f)(x_1)=(g\circ f)(x_2).$$

Because $g\circ f$ is injective, this equality implies $x_1=x_2$. That is exactly the definition of injectivity for $f$.

Another way to see the idea is that if $f$ sent two distinct inputs to the same element of $B$, no later function $g$ could separate them again. Their composite outputs would also be equal, contradicting the composite’s injectivity.

Why g is surjective

A function $g:B\to C$ is surjective if every element of $C$ is the output of at least one element of $B$. Choose any $c\in C$. Since $g\circ f$ is surjective onto $C$, there is some $a\in A$ such that

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

By the definition of composition, this says $g(f(a))=c$. The value $f(a)$ belongs to $B$, so we have found an element of $B$ that $g$ maps to $c$. Since this works for every $c\in C$, $g$ is surjective.

A finite example

Let

$$A=\{1,2\},\qquad B=\{a,b,c\},\qquad C=\{X,Y\}.$$

Define $f:A\to B$ and $g:B\to C$ by

$$f(1)=a,\quad f(2)=b,$$

and

$$g(a)=X,\quad g(b)=Y,\quad g(c)=X.$$

Then the composite sends $1$ to $X$ and $2$ to $Y$. It is a bijection from $A$ to $C$. Here, $f$ is injective but not surjective, because $c$ is not an output of $f$. Also, $g$ is surjective but not injective, because both $a$ and $c$ map to $X$. This illustrates exactly what the theorem guarantees—and what it does not.

What the theorem does not say

The composite being bijective does not imply that $f$ is surjective or that $g$ is injective. The finite example above is a counterexample to both extra claims.

The reverse implication also fails: knowing only that $f$ is injective and $g$ is surjective does not ensure that $g\circ f$ is bijective. For example, let $A=\{1\}$, $B=\{a,b\}$, and $C=\{X,Y\}$, with $f(1)=a$, $g(a)=X$, and $g(b)=Y$. The function $f$ is injective and $g$ is surjective, but the composite never reaches $Y$, so it is not bijective.

Common mistakes

• Reversing the roles: For $g\circ f$, the conclusion is that the first function $f$ is injective and the second function $g$ is surjective. The order follows from tracing inputs through the composition.
• Claiming both functions are bijective: The composite can be bijective even when $f$ misses elements of $B$ and $g$ sends multiple elements of $B$ to the same value.
• Ignoring the codomain: Surjectivity depends on the stated codomain. The proof that $g$ is onto uses the fact that the composite is onto $C$.
• Using the wrong composition: The result is about $g\circ f$, which means apply $f$ first and then $g$. The function types and conclusions must be checked against that order.