Janich Linear Algebra chapter 1 problem 1.2

Question

Let $f:X\to Y$ and $g:Y\to X$ be maps such that $fg=id_Y$ and $gf=Id_X$, then $f$ is bijective.

I already proved that $f$ is injective. Let $x,x'\in X$ and $f(x)=f(x')$. Then we have $g(f(x))=g(f(x'))$ and therefore $x=x'$. Hence $f(x)=f(x') \iff x=x'$ and $f$ is injective.

How do I prove that $f$ is surjective?

Let $y\in Y$. Choose $x=\cdots$. Then ... Therefore $f(x)=y$.

Can someone help me?

score 2 · Accepted Answer · answered May 14 '16 at 11:23

2

Set $x = g(y)$. Then $f(x) = f(g(y)) = y$.

answered May 14 '16 at 11:23

kobe

41,901

How easy. Thanks – Dr. Heinz Doofenshmirtz May 14 '16 at 11:33

Janich Linear Algebra chapter 1 problem 1.2

1 Answers1