Let $\mbox{Id}_A$ be the identity mapping from a set $A$ onto itself.

Suppose I have a function $f_1:A_1\rightarrow A_2$, and that I find a function $f_2:A_2\rightarrow A_1$ such that $f_2\circ f_1=\mbox{Id}_{A_1}$ and $f_1\circ f_2=\mbox{Id}_{A_2}$. Is this enough to prove that $f_1$ is bijective, with inverse $f_1^{-1}=f_2$?

If not, what are some simple counterexamples?


1 Answer 1


Surjectivity: Given $y \in A_2$, let $x = f^{-1}(y)$. Then $f(x)=f(f^{-1}(y))=y$. Q.E.D.
Injectivity: $f(a)=f(b) \implies f^{-1}(f(a))=f^{-1}(f(b))\implies a=b$. Q.E.D.
Bijectivity: Follows from surjectivity and bijectivity.

Alternatively, you could use the existence of $f^{-1}$ as the definition of bijectivity and then prove that surjectivity and injectivity implies bijectivity.


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