Let $A$, $B$ & $C$ sets, and left $f:A \rightarrow B$ and $g:B \rightarrow C$ be functions. Suppose that $f$ and $g$ have inverses. Prove that $g\circ f$ has an inverse, and that $(g\circ f)^{−1} = f^{−1}\circ g^{−1}$.
Assuming that f and g have reverse,$f^{-1}=h$ and $g^{-1}=s$ with $h:B\rightarrow A$ e $s:C\rightarrow B$.
from that above i infer that the inverse of $(g \circ f)$ is
$(s \circ g):C\rightarrow A$ that is $g^{-1} \circ f^{-1}=(g \circ f)^{-1}$;
Hence for proof of $(g \circ f)^{-1}=f^{-1} \circ g^{-1}$, proceed as before, only swapping functions , right?
Best Answer
Here is the correct proof. $$(g \circ f) \circ (f^{-1} \circ g^{-1}) = (g \circ (f \circ f^{-1})) \circ g^{-1} = (g \circ id_B) \circ g^{-1} =g \circ g^{-1} =id_A$$ and $$(f^{-1} \circ g^{-1}) \circ (g \circ f) = \mbox{ same computations } = id_C$$ so $(f^{-1} \circ g^{-1})$ and $(g \circ f)$ are inverses each other.