## Related questions with answers

Suppose $A$ and $B$ are sets, and $f$ and $g$ are functions with $f:A\rightarrow B$ and $g:B\rightarrow A$. Prove: If $g\circ f=\text{id}_A$ and $f\circ g=\text{id}_B$, then $f$ is invertible and $g=f^{-1}$. Note: This result is a converse to Proposition 26.9.

Solution

Verified$f : A \rightarrow B$ and $g :B \rightarrow A$ are such that $g \circ f=id_{A}$ and $f \circ g=id_{B}$.

To show: $f$ is invertible i.e it is bijective.

For $x,y \in A$ if $f(x)=f(y)$ then $g(f(x))=g(f(y))$ which implies $id_{A}(x)=id_{A}(y)$ which implies $x=y$. Therefore $f$ is one-to-one.

For any $b \in B$, $id_{B}(b)=b$ which implies $f \circ g(b)=b$ which implies $f(g(b))=b$ which implies that $f$ is onto function.

So, $f$ is a bijective and therefore an invertible function.

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