# Thread: Inverse Composition of Functions Proof

1. ## Inverse Composition of Functions Proof

I'm new here, though I wish I had found this forum long ago. (I'm an applied math major.)

I've run into trouble on my homework which is, of course, due tomorrow. Here we go:

If f: A -> B and g: B -> C are one-to-one functions, show that (g o f)^-1 = f^-1 o g^-1 on Range (g o f).

I'm stuck on where to even start on this proof, and it's driving me mad. For the record, the above notation ^-1 denotes "inverse". Please help?

2. Originally Posted by Jedi Arashi
I'm new here, though I wish I had found this forum long ago. (I'm an applied math major.)

I've run into trouble on my homework which is, of course, due tomorrow. Here we go:

If f: A -> B and g: B -> C are one-to-one functions, show that (g o f)^-1 = f^-1 o g^-1 on Range (g o f).

I'm stuck on where to even start on this proof, and it's driving me mad. For the record, the above notation ^-1 denotes "inverse". Please help?
You need to show that $(f^{-1}\circ g^{-1})\circ (g\circ f)= (g\circ f)\circ (f^{-1}\circ g^{-1}) = 1$ by 1, it means identity. Because the definition of $(g\circ f)^{-1}$ is a function which composed (either way) with $g\circ f$ gives us the identity.

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

I'm not sure I'm on the same page with you here with the above. Let me see if I can use this math code to produce what I'm saying, because I feel like I'm not quite getting it still.

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

Okay, that's what I'm supposed to prove. Do I still do the same as above, or does the strategy change?

4. Use the following proposition:The functions $f: X \to Y$ and $g: Y \to X$ are inverses of each other if and only if $g \circ f = I_X$ and $f \circ g = I_Y$.

5. Originally Posted by tukeywilliams
Use the following proposition:The functions $f: X \to Y$ and $g: Y \to X$ are inverses of each other if and only if $g \circ f = I_X$ and $f \circ g = I_Y$.
But I'm only given that f is one-to-one, and g is one-to-one (which means $g \circ f$ is one-to-one as well). They aren't inverses of one another.

6. Its an if and only if statement. $f: X \to Y$ and $g: Y \to X$ are inverses of each other $\Leftrightarrow g \circ f = I_X$ and $f \circ g = I_Y$.

So start by taking $(g \circ f) \circ (f^{-1} \circ g^{-1})$ and $(f^{-1} \circ g^{-1}) \circ (g \circ f)$.

7. It's $f: A \to B$ and $g: B \to C$. Or, I suppose if you prefer, $f: X \to Y$ and $g: Y \to Z$.

You'll have to pardon me if I sound rather confused (well, I am, forgive me!). I've spent the past hour or so at least on this problem, and proofs have never been anything near my strong point.

Crud, I'm sorry! I mixed up the latter equation!

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

I'm such an idiot, I'm sorry! That's what I'm supposed to prove.

8. First show that $g \circ f$ is a bijection and then show that its inverse function is $f^{-1} \circ g^{-1}$ using the above proposition.

9. I only have to prove it within the range of $g \circ f$. Is it still required that I prove it to be a bijection, or is there another way to go about it?

10. Recall that functions are just relations (sets of ordered pairs) with special restrictions.
In general this property is true for relations:
$\left( {p,q} \right) \in g \circ f\; \Leftrightarrow \;\left( {\exists x} \right)\left[ {\left( {p,x} \right) \in f \wedge \left( {x,q} \right) \in g} \right]$
This means that:
$\left[ {\left( {p,x} \right) \in f \wedge \left( {x,q} \right) \in g} \right]\; \Leftrightarrow \;\left[ {\left( {x,p} \right) \in f^{ - 1} \wedge \left( {q,x} \right) \in g^{ - 1} } \right]\; \Leftrightarrow \;\left( {q,p} \right) \in f^{ - 1} \circ g^{ - 1}$

Because you are dealing with only the ranges of the two functions, assume they are bijections.