Register to reply

Invert a triple composite function p(q(r(x)))

by infk
Tags: composite, function, invert, pqrx, triple
Share this thread:
infk
#1
Nov19-12, 05:41 PM
P: 21
Hey,
Let ##(f,g) \in B^A## where ##A## and ##B## are non-empty sets, ##B^A## denotes the set of bijective functions between ##A## and ##B##.
We assume that there exists ##h_0: A \rightarrow A## and ##h_1: B \rightarrow B## such that ##f = h_1 \circ g \circ h_0 ##.
This implies that ##g = h^{-1}_1 \circ f \circ h^{-1}_0##, according to my teacher, but why is that?
We have that ##h^{-1}_1 \circ f = g \circ h^{-1}_0 ##, but how do I proceed from that?

I have drawn a graph with sets and arrows representing functions such that going from ##A## to ##B## via ##h_0##, ##g## and ##h_1## is the same as going from ##A## to ##B## via ##f##. I then manipulated this graph a little while maintaining the proper relations between the sets, which showed that going from ##A## to ##B## via ##g## is the same as going through (in order) ##h^{-1}_0##, ##f## and ##h^{-1}_1##, but this is hardly a proof at all.

Thanks
Phys.Org News Partner Science news on Phys.org
Suddenly, the sun is eerily quiet: Where did the sunspots go?
'Moral victories' might spare you from losing again
Mammoth and mastodon behavior was less roam, more stay at home
Stephen Tashi
#2
Nov20-12, 11:28 AM
Sci Advisor
P: 3,245
Quote Quote by infk View Post
We have that ##h^{-1}_1 \circ f = g \circ h^{-1}_0 ##, but how do I proceed from that?
Using magic and superstition, if we begin with the equation [itex] h(x) = h(x) [/itex] and apply "equal functions" to both sides of that equation then we can get to [itex] h^{-1}( f ( h(x)) = g(h^{-1}(h(x)) = g(x) [/itex]. The problem is how to justify that procedure.


Your materials may have proven that procedure as a theorem. We'd need to see those materials in order to find an easy proof. Are you asking for some "slick" way of proving the result?

A knockdown-drag-out roof could begin: "Consider the two functions [itex] A(x) = h^{-1}f( h(x)) [/itex] and [itex] B(x) = g(h^{-1}(h(x)) [/itex] ."

If [itex] A(x) = y [/itex] then show that [itex] x = h((f^{-1}(h^{-1}(x)) [/itex]. Use that expression for [itex] x [/itex] to show that [itex] B(x) = y [/itex] also. In a similar manner show that if [itex] B(x) = y [/itex] then [itex] A(x) = y [/itex]. Argue that [itex] A(x) [/itex] and [itex] B(x) [/itex] have the same domain and range. Thus they are identical functions.

.
infk
#3
Nov21-12, 05:50 AM
P: 21
Quote Quote by Stephen Tashi View Post
Using magic and superstition, if we begin with the equation [itex] h(x) = h(x) [/itex] and apply "equal functions" to both sides of that equation then we can get to [itex] h^{-1}( f ( h(x)) = g(h^{-1}(h(x)) = g(x) [/itex]. The problem is how to justify that procedure.


Your materials may have proven that procedure as a theorem. We'd need to see those materials in order to find an easy proof. Are you asking for some "slick" way of proving the result?

A knockdown-drag-out roof could begin: "Consider the two functions [itex] A(x) = h^{-1}f( h(x)) [/itex] and [itex] B(x) = g(h^{-1}(h(x)) [/itex] ."

If [itex] A(x) = y [/itex] then show that [itex] x = h((f^{-1}(h^{-1}(x)) [/itex]. Use that expression for [itex] x [/itex] to show that [itex] B(x) = y [/itex] also. In a similar manner show that if [itex] B(x) = y [/itex] then [itex] A(x) = y [/itex]. Argue that [itex] A(x) [/itex] and [itex] B(x) [/itex] have the same domain and range. Thus they are identical functions.

.
Hi and thanks for the response. What I meant was of course that ##h^{-1}_1 \circ f = g \circ h_0##, it seems though that you did not notice my typo. Also, could you use subscript notation, it is not clear which of ##h_1## and ##h_2## you mean. Cheers

Stephen Tashi
#4
Nov23-12, 07:42 PM
Sci Advisor
P: 3,245
Invert a triple composite function p(q(r(x)))

Using magic and superstition, if we begin with the equation [itex] h_0^{-1}(x) = h_0^{-1}(x) [/itex] and apply "equal functions" to both sides of that equation then we can get to [itex] h_1^{-1}( f ( h_0^{-1}(x)) = g(h_0(h_0^{-1}(x)) = g(x) [/itex]. The problem is how to justify that procedure.

Your materials may have proven that procedure as a theorem. We'd need to see those materials in order to find an easy proof. Are you asking for some "slick" way of proving the result?


Register to reply

Related Discussions
Composite function Calculus & Beyond Homework 1
Composite function Set Theory, Logic, Probability, Statistics 4
Composite Function Precalculus Mathematics Homework 3
Onto composite function Calculus & Beyond Homework 4
Composite and one to one function Calculus & Beyond Homework 2