Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am stuck with this question,

Let $A=B=C=\mathbb{R}$ and consider the functions $f\colon A\to B$ and $g\colon B\to C$ defined by $f(a)=2a+1$, $g(b)=b/3$. Verify Theorem 3(b): $(g\circ f)^{-1}=f^{-1}\circ g^{-1}.$

I have calculated $f^{-1}$, $g^{-1}$, and their composition, but how do I find the inverse of $(g\circ f)$?

Here is how I have done so far,

$$\begin{align*} \text{Let}\qquad\qquad b &= f(a)\\ a&= f^{-1}(b)\\ &{ }\\ b&=f(a)\\ b&=2a+1\\ \frac{b-1}{2} &= a\\ a &= \frac{b-1}{2} \end{align*}$$ But $a=f^{-1}(b)$, $$f^{-1}(b) = \frac{b-1}{2}.$$


$$\begin{align*} \text{Let}\qquad\qquad a&=g(b)\\ b&= g^{-1}(a)\\ a&= g(b)\\ a &= b/3\\ b &= 3a\\ g^{-1}(a) &= 3a\qquad(\text{because }b=g^{-1}(a) \end{align*}$$


$$\begin{align*} f^{-1}\circ g^{-1} &= ?\\ &= f^{-1}\Bigl( g^{-1}(a)\Bigr)\\ &= f^{-1}(3a)\\ f^{-1}\circ g^{-1} &= \frac{3a-1}{2} \end{align*}$$

$$\begin{align*} g\circ f&= g\bigl(f(a)\bigr)\\ &= g(2a+1)\\ g\circ f &= \frac{2a+1}{3}\\ (g\circ f)^{-1} &= ?\\ \text{Let}\qquad\qquad &b=g\circ f \end{align*}$$


Thanks for the answers, I followed the suggestions and came up with the answer, enter image description here enter image description here Now I have two questions,

  1. The answers do match but the arguments are different. Is that ok?
  2. Is $(g\circ f)$ same as $(g\circ f(a))$?
share|improve this question
$(g\circ f)$ is a function. $(g\circ f(a))$ is the value of the function $g\circ f$ at $a$. They are not the same thing (one is a function, the other is an number that you've written in parentheses). The name of the variable doesn't matter. The function $g(x)=x^2$ is the same as the function $g(z)=z^2$. –  Arturo Magidin Aug 26 '11 at 16:46
add comment

3 Answers

up vote 3 down vote accepted

$f,g$ are the functions defined in the question.

enter image description here

We have

$$\begin{eqnarray*} b &=&f(a)=2a+1, \end{eqnarray*}$$

or equivalently, by definition of the inverse function $f^{-1}$

$$\begin{eqnarray*} &a=\frac{b-1}{2}=f^{-1}(b).\tag{A} \end{eqnarray*}$$


$$\begin{eqnarray*} c &=&g(b)=\frac{b}{3}, \end{eqnarray*}$$

or equivalently, by definition of the inverse function $g^{-1}$

$$\begin{eqnarray*} b=3c=g^{-1}(c),\tag{B} \end{eqnarray*}$$

after combining $(A)$ and $(B)$, we get

$$a=\frac{3c-1}{2}=(f^{-1}\circ g^{-1})(c).\tag{1}$$

On the other hand

$$c=(g\circ f)(a)=g(f(a))=g(2a+1)=\frac{2a+1}{3}.\tag{2}$$

Hence, by definition, the value at $c$ of the inverse function $(g\circ f)^{-1}$, is

$$a=\frac{3c-1}{2}=(g\circ f)^{-1}(c).\tag{3}$$

From $(1)$ and $(3)$ we conclude that for these functions $f,g$ and their inverses $f^{-1},g^{-1}$ the following identity holds:

$$(f^{-1}\circ g^{-1})(c)=(g\circ f)^{-1}(c).\tag{4}$$

Notation's note: $(f^{-1}\circ g^{-1})(c)=f^{-1}(g^{-1}(c))$.

share|improve this answer
Thanks a lot for such an effort. –  Fahad Uddin Aug 26 '11 at 19:25
@fahad: You are welcome. –  Américo Tavares Aug 26 '11 at 19:28
@AméricoTavares I usually draw the function in this way eevry time it helps me with composition of functions and when I "play" with algebraic structures but I was always scared to use them in order to explain my concepts in my questions to other people because I believed I was taking too much freedom with a notation that I only saw in cathegory theory, where your sets A,B and C are usually set with sturctures. So is correct to use these "diagrams" even outside the cathegory theory context? –  MphLee Apr 20 '13 at 16:25
@MphLee I am not a mathematician and know nothing about cathegory theory, but these diagrams appeared in some books of Calculus/Real Analysis for Engineers back in 1060-1970's. –  Américo Tavares Apr 20 '13 at 17:03
@AméricoTavares and I'm even less a mathematican I'll try to search more about this and I'll continue use these diagrams in my amatorial practice, thanks anyways –  MphLee Apr 20 '13 at 17:44
add comment

What you've done so far is to compute $f^{-1}$ and $g^{-1}$, and $f^{-1}\circ g^{-1}$. Now you want to try to find $(g\circ f)^{-1}$ directly, and compare that to what you've computed (in order to verify the formula).

So, you've figured out that $(g\circ f)(a) = \frac{2a+1}{3}$. How do we figure out $(g\circ f)^{-1}$?

Exactly the same way we figure out the inverse of any function. If someone stopped you on the street, pointed a gun at you and said

"Here, I have this function: $$h(a) = \frac{2a+1}{3},$$ I need the formula for $h^{-1}$. Give it to me or I'll shoot you!"

then you don't need to know where that function came from, all you need to do is figure out the inverse: $$\begin{align*} b &= \frac{2a+1}{3}\\ 3b &= 2a+1\\ 3b-1 &= 2a\\ &\vdots \end{align*}$$ etc. When you are done and have a formula for $h^{-1}(a) = (g\circ f)^{-1}(a)$, you can compare it to the formula you found for $f^{-1}\circ g^{-1}$ and verify that you got the same function.

share|improve this answer
Thanks alot. I did it as you suggested. A small problem that I am having is that is in terms of b instead of a. Checkout the edit. –  Fahad Uddin Aug 26 '11 at 16:30
@fahad: Don't get hung up on the letters. The name of the variable is immaterial. The function $h(x) = x^2$ is exactly the same as the function $h(y)=y^2$, which is exactly the same as the function $h(z)=z^2$, which is exactly the same as the function $h(a)=a^2$. Just switch all the $b$s into $a$s and be done. –  Arturo Magidin Aug 26 '11 at 16:36
Thanks alot for the answer :) –  Fahad Uddin Aug 26 '11 at 19:15
I can't resist adding that the Russian mathematical physicist Igor Tamm was once told to do a mathematics calculation or be shot. :) –  Mike Spivey Aug 26 '11 at 22:21
add comment

You find the inverse of $g\circ f$ by using the fact that $(g\circ f)^{-1} = f^{-1} \circ g^{-1}$.

In other words, what gets done last gets undone first.

$f$ multiplies by 2 and then adds 1.

$g$ divides by 3.

Dividing by 3 is done last, so it's undone first.

The inverse first multiplies by 3, then undoes $f$.

Later note: Per the comment, to verify that $(g\circ f)^{-1} = f^{-1} \circ g^{-1}$:

Instead of confusingly writing $a = g(b)$, write $c=g(b)$. Then $c=b/3$, so $b=3c$, so $$g^{-1}(c) = 3c.$$ And $$f^{-1}(b) = \frac{b-1}{2}.$$

So $$ b = 3c\qquad\text{and}\qquad a = \frac{b-1}{2}. $$ Put $3c$ where $b$ is and get $$ a=\frac{3c-1}{2}. $$ You want to show that that's the same as what you'd get by finding $g(f(a))$ directly and then inverting.

So $c = g(f(a)) = \dfrac{f(a)}{3} = \dfrac{2a+1}{3}$.

So take $c = \dfrac{2a+1}{3}$ and solve it for $a$:

$$ \begin{align} 3c & = 2a+1 \\ 3c - 1 & = 2a \\ \\ \frac{3c-1}{2} & = a. \end{align} $$

FINALLY, observe that you got the same thing both ways.

share|improve this answer
But the problem asks the student to verify the formula; that is, find the inverse of $g\circ f$ "directly", and then compare it to the function you get by computing $f^{-1}\circ g^{-1}$. Surely using the formula to verify that the formula works is a tad... unsatisfying. –  Arturo Magidin Aug 26 '11 at 15:58
I am stuck with how do I find (gof)^-1 –  Fahad Uddin Aug 26 '11 at 16:14
OK, I've added a later note. –  Michael Hardy Aug 26 '11 at 16:18
@MichaelHardy thank you for this answer, it came in handy for a question I recently asked –  Assad Dec 15 '13 at 13:01
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.