Proving how to apply the adjoint to a dyad I have to prove that  $(|\Psi\rangle \langle \Phi|)^\dagger= |\Phi\rangle \langle \Psi|$
Hint: Prove that $\langle f|(|\Psi\rangle \langle \Phi|)^\dagger|g\rangle = ... =  \langle f||\Phi\rangle \langle \Psi||g\rangle$
 A: The notation obsecures the meaning but I'll try to cope with it by writing the definition of the adjoint of $A$ as
$$
\langle f|A|g\rangle = \overline{\langle g|A^\dagger|f\rangle}\\
$$
The given operator is in a tensor (dyad) form. It is to be understood as scaling the lefthand ket by the inner product of the right bra with the argument, (on which the operator is applied)
$$
\langle f||\Phi\rangle \langle \Psi||g\rangle = \overline{\langle \Phi||f\rangle \langle g||\Psi\rangle}
= \overline{ \langle g||\Psi\rangle \langle \Phi||f\rangle} 
=  \overline{ \langle g|(|\Phi\rangle \langle\Psi |)^\dagger|f\rangle} 
$$
In a more clear notation and for better explanation we do it without the Dirac notation but the meaning of the operator is as above
$$
A = |\Phi\rangle \langle \Psi| = \Psi\langle\Phi,\cdot\rangle
$$
this acts on $x$ like
$$
Ax = \Psi\langle\Phi,x\rangle
$$
The Definition of the adjoint of $A$ is
$$
\langle f, Ag \rangle = \langle A^\dagger f, g \rangle
$$
Proving the statement
$$
\langle f, Ag \rangle 
= \langle f, \Psi\langle\Phi,g\rangle \rangle 
= \langle  f, \Psi \rangle \langle\Phi,g\rangle
=  \langle \overline{\langle  f, \Psi \rangle} \Phi,g\rangle
=  \langle \langle  \Psi, f \rangle \Phi,g\rangle
=  \langle\Phi \langle  \Psi, f \rangle ,g\rangle
=  \langle A^\dagger f  ,g\rangle
$$
Where
$$
A^\dagger = \Phi \langle  \Psi, \cdot \rangle = |\Psi\rangle \langle \Phi|
$$
A: Let $A = |\Psi\rangle \langle \Phi|$, then
$$
\langle f|A^\dagger|g\rangle = \langle g|A|f\rangle^* = \big(\langle g|\Psi\rangle \langle \Phi|f\rangle\big)^* =  \langle g|\Psi\rangle^* \langle \Phi|f\rangle^* = \langle\Psi|g\rangle\langle f|\Phi\rangle = \langle f|\Phi\rangle \langle\Psi|g\rangle,
$$
so $A^\dagger = |\Phi\rangle \langle\Psi|$.
