0 like 0 dislike
155 views
Proof: Let $f: A \rightarrow B$ and $g: B \rightarrow C$ be functions. If $g \circ f$ is bijective, then $f$ is injective and $g$ is surjective.
| 155 views

0 like 0 dislike
Proof. Suppose the statement does not hold, so $f$ is not injective or $g$ is not surjective. Let us consider both cases:
- $\mathrm{f}$ is not injective, which means that $\exists a, a^{\prime} \in A$ such that $f(a)=f\left(a^{\prime}\right)$.
Now,
$(g \circ f)(a)=g(f(a))=g\left(f\left(a^{\prime}\right)\right)$
so we have $(g \circ f)(a)=(g \circ f)\left(a^{\prime}\right)$ but $a \neq a^{\prime}$. So $g \circ f$ is not injective, hence it is not bijective.
- g is not surjective, which means that $\exists c \in C$ such that for all $b \in B, g(b) \neq c .$ Moreover, $g \circ f$ is surjective, so $\exists a \in A$ such that $(g \circ f)(a)=c .$ Now, if $b=f(a)$, then $g(b)=c$, which is a contradiction!

Both cases lead us to the contradiction, hence we may conclude that if $g \circ f$ is bijective, then $f$ is injective and $g$ is surjective.
by Platinum (130,522 points)

0 like 0 dislike
1 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike