I've been trying for a while now to prove that $(g \circ f) = ((g\circ f)^{-1})^{-1}$ as a corollary to one of the exercises in Tao's Analysis I book. I am trying to restrict myself to the following strategy because it highlights an issue that I have seen come up several time in my proofs.
The goal of this proof is to conclude with the statement that $\forall x \in X, (g \circ f (x)) = (((g\circ f)^{-1})^{-1}(x))$. This is the definition of function equality.
Also, just for clarity: $f: X \to Y$ and $g:Y \to Z$
Here is what I know from prior exercises and prior proofs in the book:
- $(g\circ f)$ is bijective
- $(g \circ f)^{-1}$ is bijective
- $((g \circ f)^{-1})^{-1}$ is bijective
- $(g \circ f)^{-1}(g \circ f \ (x)) = x$
- $(g \circ f)^{-1}(((g \circ f)^{-1})^{-1}\ (x))=x$
The proof is as follows:
From $4$ and $5$ we know that: $(g \circ f)^{-1}(g \circ f \ (x)) = (g \circ f)^{-1}(((g \circ f)^{-1})^{-1}\ (x))$
From $2$, we have the following statement: $\forall z \in Z$ if $(g \circ f)^{-1}(z)=(g \circ f)^{-1}(z')$, then $z=z'$.
From $1$ and $3$ we have: $(g \circ f \ (X))=Z$ and $(((g \circ f)^{-1})^{-1} \ (X))=Z$.
Because every $z^* \in Z$ can be represented by a particular $(g \circ f \ (x^*))$ and a particular $(((g \circ f)^{-1})^{-1} \ (x^*))$, I am going to rewrite my earlier statement as:
$\forall z \in Z$ if $(g \circ f)^{-1}(g \circ f \ (x))=(g \circ f)^{-1}(((g \circ f)^{-1})^{-1}\ (x))$, then $(g \circ f \ (x))=(((g \circ f)^{-1})^{-1}\ (x))$.
Now, the consequent of this statement is certainly part of what I want...however, I cannot figure out how (informally stated) $\forall z \in Z \implies \forall x \in X$. I am struggling with the logical step(s) to get to this concluding statement. It's certainly intuitive that this SHOULD BE the case, but I cannot for the life of me figure out the proper wording to get here.
Any help is greatly appreciated!