Q:

Isomorphism and its effects on function composition

In a proof the professor did this while talking about the transformation matrix 
$$ \Phi_{A'} \circ \tilde{T}_{A'}^{A} = id_V \circ \Phi_{A} = \Phi_{A} \rightarrow \quad \tilde{T}_{A'}^{A} = \Phi_{A'}^{-1}  \circ \Phi_{A}$$
He basically just put $\Phi_{A'}$ on the other side, much like an algebraic transformation and said he was allowed to do it, because all the linear mappings above are isomorphisms. 
Could somebody elaborate, why isomorphisms allow you do that?

A:

Since $\Phi_{A'}$ is an isomorphism, $id = \Phi^{-1}_{A'}\circ\Phi_{A'}$. Hence,
$$\tilde{T}_{A'}^A = id\circ \tilde{T}_{A'}^A = \Phi^{-1}_{A'}\circ\Phi_{A'}\circ \tilde{T}_{A'}^A \overset{!}{=} \Phi^{-1}_{A'}\circ\Phi_A.$$
The equality (!) used the fact that $\Phi_{A'}\circ\tilde{T}_{A'}^A = \Phi_A$.

