Show that $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$

Question

Let $f: A \to B$ and $g: B \to C$ be invertible mappings; that is, mappings such that $f^{-1}$ and $g^{-1}$ exist. Show that $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$.

I tried using the facts that $f^{-1}(f(x)) = x$ and $g^{-1}(g(x)) = x$ but that didn't seem to help and got me trivial statements.

apply (g o f)(x) from right on both sides to your claim, and then use associativity on the right side and the definition of inverse on the left side — Mihail, May 27 '16 at 20:23

score 5 · Answer 1 · edited Apr 06 '24 at 14:10

5

You have to show that $f^{-1}\circ g^{-1}$ is the inverse of $g\circ f$. Hence you have to show that $$f^{-1}\circ g^{-1} \circ g\circ f = id_A$$ and $$g\circ f \circ f^{-1}\circ g^{-1} = id_C.$$ Can you finish ?

edited Apr 06 '24 at 14:10

Anne Bauval

51,477

answered May 27 '16 at 20:23

C. Dubussy

9,403

So a function is invertible if it has both a pre and post inverse? – Puzzled417 May 27 '16 at 20:32
1

Yes, $g$ is the inverse of $f$ iff $g\circ f = id$ and $f\circ g = id$. – C. Dubussy May 27 '16 at 20:35

Show that $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$

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