Date modified: Sat 2024-03-23 . 01:18 AM
Date created: Fri 2024-03-22 . 02:11 AM
# 23b Uniqueness of inverse function. Here we claim the following > If a function $f:A\to B$ is invertible, then its inverse function is unique. Hence it makes sense to speak of **the** inverse of $f$. Again, recall a function $f:A\to B$ is invertible if there exists a function $g:B\to A$ such that $g\circ f=\text{id}_{A}$ and $f\circ g=\text{id}_{B}$. $\blacktriangleright$ Proof. Suppose $f:A\to B$ is invertible, and that there exists two functions $g,h:B\to A$ that are both inverses to $f$. We want to show $g=h$. To show two functions are the same, we show for each $x\in B$ we have $g(x)=h(x)$. Indeed, consider the two points $g(x)$ and $h(x)$, and apply $f$ to both sides. We get $f(g(x))=f(h(x))=x$. From previous we showed $f$ being invertible is bijective and hence injective, so we must have $g(x)=h(x)$. $\blacksquare$ Above is an element-wise proof of this fact, which we used the characterization of bijective iff invertible. There is a simpler proof using just associativity of function composition: $\blacktriangleright$ Proof, again. Suppose $f:A\to B$ is invertible, and suppose $g,h:B\to A$ are both inverses to $f$. Then $$ f\circ g=\text{id}_{B}=f\circ h. $$ Now apply $g$ to both sides, and we get $$ g\circ f \circ g=g\circ f \circ h $$But $g\circ f=\text{id}_{A}$, so we get $$ \text{id}_{A}\circ g=\text{id}_{A}\circ h. $$ Since $\text{id}_{A}(x)=x$ for all $x\in A$, we have $\text{id}_{A}\circ g=g$ and $\text{id}_{A}\circ h=h$. Hence $g=h$. $\blacksquare$