Let $f: A\rightarrow B$ and $g: B\rightarrow C$ be invertible maps, show that $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$.

Question

I am working on the following problem for my abstract algebra class, and I wanted to get some feed back to see if I am on the right track. Here is what I have so far.

Let $f: A\rightarrow B$ and $g: B\rightarrow C$ be invertible maps, show that $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$.

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

Answer 1

1) Any proof that only contains math symbols but no English (or whatever you favourite language for human communication is) is probably insufficient.

2) Are you computing something above? perhaps you want to use some $=$ symbols somewhere.

3) Start by recalling the definition of the inverse function. Then list what you know (i.e., that $f^{-1}$ is the inverse of $f$, but spell out what that means, and similarly for $g$). Then write down what you want to prove (i.e., spell out what needs to be shown to establish that $(g\circ f)^{-1}=f^{-1}\circ g^{-1}$). Now try to get from what you know to what you want to know.

The proof as it currently stands is, at beast, unreadable.

Answer 2

$$\mbox{id}= \mbox{id}\\ \mbox{id}= \mbox{id}\circ\mbox{id}\\ \mbox{id}= f^{-1}\circ f\circ\mbox{id}\\ \mbox{id}= f^{-1}\circ\mbox{id}\circ f\\ \mbox{id}= f^{-1}\circ g^{-1}\circ(g\circ f)\\ (g\circ f)^{-1}= f^{-1}\circ g^{-1}\\$$ We can conclude that $$(g\circ f)^{-1}= f^{-1}\circ g^{-1}$$ Q.E.D.

Answer 3

Let $h: C \rightarrow A=(g \circ f)^{-1}$. So, $(g \circ f)\circ h=1_C$.

Function composition is associative, so: $$g \circ(f \circ h)=1_C$$ $$f \circ h=1_B \circ (f \circ h)=(g^{-1} \circ g)\circ(f \circ h)=g^{-1} \circ (g \circ(f \circ h))=g^{-1} \circ 1_C=g^{-1}$$ So: $$(g \circ f)^{-1}=h=1_A \circ h = (f^{-1} \circ f) \circ h = f^{-1} \circ (f \circ h)=f^{-1} \circ g^{-1}$$

Answer 4

$\square^{-1}=\triangle\iff\square\mathbin\circ\triangle=\mathrm{id}$. In our case show that $(g\mathbin\circ f)\mathbin\circ(f^{-1}\mathbin\circ g^{-1})=\mathrm{id}$ using that $\circ$ is associative.