Register to reply 
Invert a triple composite function p(q(r(x))) 
Share this thread: 
#1
Nov1912, 05:41 PM

P: 21

Hey,
Let ##(f,g) \in B^A## where ##A## and ##B## are nonempty 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 


#2
Nov2012, 11:28 AM

Sci Advisor
P: 3,254

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 knockdowndragout 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. . 


#3
Nov2112, 05:50 AM

P: 21




#4
Nov2312, 07:42 PM

Sci Advisor
P: 3,254

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 