# Thread: Help on invertible function question

1. ## Help on invertible function question

I like to know if anyone can tell me if my answer is correct.

Suppose f is an invertible function from Y to Z and g is an invertible function from X to Y. Show that the inverse of the composition is f o g is given by (f o g)-1 = g-1 o f -1

f –1(z) = y, g-1(y) =x
(f o g)-1(y) = f -1 (g(y))-1 =x

(g-1 o f -1)(y) = g-1(f(y))-1= z

Therefore, (f o g)-1 = g-1 o f -1 are not equal. In other words, the commutative law does not hold for the composition function.

(The f -1 and g-1 stands for the power of -1, sorry i don't have the button function to purt it correctly)

2. $\displaystyle \begin{array}{rcl} {\left( {c,a} \right) \in \left( {f \circ g} \right)^{ - 1} } & \Leftrightarrow & {\left( {a,c} \right) \in f \circ g} \\ {} & \Rightarrow & {\left( {\exists b} \right)\left[ {\left( {a,b} \right) \in g \wedge \left( {b,c} \right) \in f} \right]} \\\end{array}$$\displaystyle \begin{array}{rcl} {} & \Rightarrow & {\left( {c,b} \right) \in f^{ - 1} \wedge \left( {b,a} \right) \in g^{ - 1} } \\ {} & \Rightarrow & {\left( {c,a} \right) \in g^{ - 1} \circ f^{ - 1} } \\ \end{array}$

Now you can follow that and do the other half.

3. i appreciate your help Plato, thanks a lot