# Math Help - Inverse and composition functions

1. ## Inverse and composition functions

I have this problem from my discrete math comp sci class about inverse functions

show that for the functions f, g and h

$(fog^-1oh)^-1 = h^-1 o g o f^-1$

All i have so far is that

$h^-1 o g o f^-1 = h^-1 o g o f^-1$

because I have in my notes that $(gof)^-1 = f^-1 o g^-1$

If that right? And if so is there any other way I can show more work?

2. Hint: to produce $f^{-1}$ and $f\circ g$, write f^{-1} and f\circ g.

All i have so far is that

$h^-1 o g o f^-1 = h^-1 o g o f^-1$
This is hardly surprising since equality is reflexive

If you know that $(f\circ g)^{-1}=g^{-1}\circ f^{-1}$ and $(f^{-1})^{-1}=f$ (e.g., when all functions are bijections), then you can write $(f\circ g^{-1}\circ h)^{-1}$, parenthesize it fully (one of the $\circ$ has to be done first; does it matter which?) and then apply the equalities above to gradually rewrite this into $h^{-1}\circ g\circ f^{-1}$ (I believe there are three steps).

3. Originally Posted by guyonfire89
I have this problem from my discrete math comp sci class about inverse functions

show that for the functions f, g and h

$(fog^-1oh)^-1 = h^-1 o g o f^-1$

All i have so far is that

$h^-1 o g o f^-1 = h^-1 o g o f^-1$

because I have in my notes that $(gof)^-1 = f^-1 o g^-1$

If that right? And if so is there any other way I can show more work?
You are not asked to derive it or anything hard, you are just asked to show' it. Therefore, just multiply them together. If you get the constant map then clearly it is your inverse. You do, so it is.

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

(Also, in LaTeX (math mode') write f^{-1} to make sure that the whole -1 is put up there. Use \circ instead of `o' for composition too.)

4. "inverses" are [b]defined[\b] by "f(x)= y, if and only if $f^{-1}(y)= x$".

Suppose that fg^{-1}h(x)= f(g^{-1}h)(x)= y. Then f^{-1}(y)= g^{-1}(h(x)) where I have taken "x" in the definition to by g^{-1}h(x). from g^{-1}(h(x))= f^{-1}(y), now with "x" on the right side of the definition taken as f^{-1}(y) and "y" as h(x), we have g(f^{-1}(y))= h(x). Finally, taking g(f^{-1}(y)) as "y", x= h^{-1}(g(f^{-1}(y))= (h^{-1}gf^{-1})(y)