# Proof

• Let $x,y\in A$ with $x\neq y$.
• By hypothesis, the composition $g\circ f$ is injective.
• Thus, $g(f(x))\neq g(f(y))$.
• Since $g$ is a function, $g$ is right-unique.
• Therefore, $f(x)\neq f(y)$, otherwise we would have $g(f(x)) = g(f(y))$, which is not the case.

• By hypothesis, the composition $g\circ f$ is surjective.
• This means that there is an element $x\in A$ with $g(f(x))=z$ for every $z\in C.$
• Set $y:=f(x)$. Since $y\in B$, there is an element $y\in B$ with $g(y)=z$ for every $z\in C.$
• Therefore, $g$ is surjective.

• By hypothesis, $f$ is surjective and $g\circ f$ is injective.
• Let $x,y\in B$ with $x\neq y$.
• Since $f$ is surjective, there are elements $u,w\in A$ with $f(u)=x$ and $f(w)=y.$
• Therefore, since $x\neq y$, we have $f(u)\neq f(w).$
• Since $g\circ f$ is injective, we have $g(f(u))\neq g(f(w)).$
• Altogether, it follows $g(x)\neq g(y).$
• Thus, $g$ is injective.

• By hypothesis, $g\circ f$ is surjective and $g$ is injective.
• Since $g\circ f$ is surjective, there is an element $x\in A$ with $g(f(x))=y$ for every $y\in B.$
• Since $g$ is injective, we have $f(x)=y.$
• In other words, there is an element $x\in A$ with $f(x)=y$ for every $y\in B.$
• Thus, $f$ is surjective.

Thank you to the contributors under CC BY-SA 4.0!

Github:

### References

#### Bibliography

1. Reinhardt F., Soeder H.: "dtv-Atlas zur Mathematik", Deutsche Taschenbuch Verlag, 1994, 10th Edition
2. Wille, D; Holz, M: "Repetitorium der Linearen Algebra", Binomi Verlag, 1994