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.

Please explain me proof of "Associative Property for Composition of Functions"

share|improve this question

closed as off-topic by Jonas Meyer, Ittay Weiss, Grigory M, Najib Idrissi, dustin Jan 18 at 18:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jonas Meyer, Ittay Weiss, Grigory M, Najib Idrissi, dustin
If this question can be reworded to fit the rules in the help center, please edit the question.

4  
@JackManey: There is no need to be rude to new users even if the question is not ideal. I think we should be more welcoming of new users. Specifically, see this meta question, and the most upvoted answer there: What should we do with poorly posed questions? –  Eric Naslund Aug 30 '12 at 23:12
1  
@EricNaslund - This question is not poorly posed--at least not in the sense that it's easy to understand what the OP is looking for--but the OP is clearly asking for help without putting forth any effort. That's not acceptable on Stack Overflow; why should it be acceptable here? –  user5137 Aug 31 '12 at 2:51
1  
@EricNaslund There is something about "Explain the answer of my question before my test day after tomorrow." that makes them easy to downvote :( –  rschwieb Aug 31 '12 at 16:31

3 Answers 3

Let $A$, $B$, $C$, and $D$ be sets. Let $f : A \rightarrow B$, $g : B \rightarrow C$, and $h : C \rightarrow D$. Take any $a \in A$.

$(h \circ (g \circ f))(a) = h ((g \circ f)(a)) = h(g(f(a)))$ $\color{red}{(1)}$

On the other hand

$((h \circ g)\circ f))(a) = (h \circ g)(f(a)) = h(g(f(a))$ $\color{blue}{(2)}$

$\color{red}{(1)}$ and $\color{blue}{(2)}$ are equal for any abitrary $a \in A$. Hence $(h \circ g)\circ f = h \circ (g \circ f)$. This gives associativity.

share|improve this answer

Both $f\circ(g\circ h)$ and $(f\circ g)\circ h$ are functions which take an argument $x$ and yield $f(g(h(x))$. Since they always yield the same result for the same argument, they are the same function.

share|improve this answer

It means that if you have three functions $f,g,h$, then $f\circ(g\circ h)$ is the same function as $(f\circ g)\circ h$.

Proving this is just a routine verification, since $(f\circ g)(x)$ is defined to be $f(g(x))$.

Let's write some of it out: $f\circ(g\circ h)(x)=f((g\circ h)(x))=f(g(h(x)))=$?

Perhaps at this point you might start at the other end ($(f\circ g)\circ h$) and see if that connects with this.

share|improve this answer

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