Does AA^\dagger=I Imply A^\dagger is the Generalized Inverse of A?

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Hi,
Does the equation [tex]AA^\dagger=I[/tex] force [tex]A^\dagger[/tex] to be the generalized inverse of A? That is: [tex]AA^\dagger=I\Rightarrow A^\dagger\text{ is the generalized inverse of } A?[/tex] A is any rectangular matrix over the field of complex numbers. It is very easy to verify the first three properties, but I'm not sure about the last one
 
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It looks the answer is yes. The result can be found in http://en.wikipedia.org/wiki/Proofs_involving_the_Moore%E2%80%93Penrose_pseudoinverse" .
 
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Hi!
I think so. Actually i think we can write out what A† is.
It is A*(AA*)^(-1), and the 4th condition follows.
 
No.
In my question A\dagger was conjugate transpose and conditions 3 and 4 in Penrose's definition are just the requirement that AX and XA both should be hermitian, where X is the generalized inverse of A (in my case which is A\dagger and so 3-4 follows trivially).