Show functions are injective, surjective if they satisfy identity function

Question

5. (c) Show that if functions $f:A→B$ and $g:B→A$ satisfy $g◦f=1_A$, where $1_A:A→A$ is the identity function of $A$, then the function $f$ is injective, while $g$ is surjective.

How will I show this if I don't even know the functions?

Answer 1

Injectivity: Let $x,y\in A$ and $f(x)=f(y)$ then $g(f(x))=g(f(y))$ since $g$ is also a function. By assumption $g(f(x))=x$ for all $x\in A$ follows $x=y$ hence $f$ is injective.

Surjectivity: We need to show that for every $x\in A$ there is some $z\in B$ such that $g(z)=x$. By definition of $f$ we have $f(x)\in B$. Take $z=f(x)$ then $g(z)=g(f(x))=x$. Hence $g$ is surjective.