Does $ \mu * N $ imply $ \mu^{-1} * N^{-1} $ where $ * $ denotes Dirichlet multiplication?

Question

Quoting from the book Introduction to Analytic Number Theory by Tom A. Apostol > Section 2.11 (page 37).

EXAMPLE The inverse of Euler's $ \varphi $ function. Since $ \varphi = \mu * N $ we have $ \varphi^{-1} = \mu^{-1} * N^{-1} $.

How does $ \varphi = \mu * N $ imply $ \varphi^{-1} = \mu^{-1} * N^{-1}$?

Note:

• The asterisk symbol $ * $ denotes Dirichlet multiplication (also known as Dirichlet convolution).
• The function $ \varphi(n) $ denotes Euler's totient function.
• The function $ \mu(n) $ denotes Mobius function.
• The function $ N(n) $ is defined as $ N(n) = n $.

I am unable to find anything in the preceding text that justifies this step. Maybe I missed something? Can someone help me to understand why this step works?