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Invert a triple composite function p(q(r(x))) 
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#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: 3,285

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: 3,285

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? 


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