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

infk

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

Stephen Tashi

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.

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infk

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

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?