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]

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 $h(x) = h(x)$ and apply "equal functions" to both sides of that equation then we can get to $h^{-1}( f ( h(x)) = g(h^{-1}(h(x)) = g(x)$. 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 $A(x) = h^{-1}f( h(x))$ and $B(x) = g(h^{-1}(h(x))$ ."

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

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[infk]

Using magic and superstition, if we begin with the equation $h(x) = h(x)$ and apply "equal functions" to both sides of that equation then we can get to $h^{-1}( f ( h(x)) = g(h^{-1}(h(x)) = g(x)$. 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 $A(x) = h^{-1}f( h(x))$ and $B(x) = g(h^{-1}(h(x))$ ."

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

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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 $h_0^{-1}(x) = h_0^{-1}(x)$ and apply "equal functions" to both sides of that equation then we can get to $h_1^{-1}( f ( h_0^{-1}(x)) = g(h_0(h_0^{-1}(x)) = g(x)$. 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?