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I want to prove that for every $n \in \mathbb{N}$ : $$$$ $f(n)=\sum_{d|n} g(d) \quad$ $ \Leftrightarrow$ $\quad g(n)=\sum_{d|n} \mu(d)f(n/d)$ $$$$ I proved the first direction and for the second one I tried the following:$$$$IC Let $$ g(n)=\sum_{d|n}\mu(d)f(n/d) $$ $$ g(n)=\sum_{d|n}\mu(n/d)f(d) $$ Now applying the first direction yields

$$\mu(n/n)f(n)=\sum_{d|n}\mu(d)g(n/d) $$

$$\mu(1)f(n)=\sum_{d|n}\mu(n/d)g(d) $$

$$f(n)=\sum_{d|n}\mu(n/d)g(d) $$ Now letting $n=2$ then $f(2) = g(2)-g(1)$ not $f(2) = g(2)+g(1)$. $$$$ Where did I go wrong? Thanks in advance.

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    $\mu(n/n)f(n)=\sum_{d|n}\mu(d)g(n/d)$ is not correct, it is $\mu(n/n)f(n)=f(n) =\sum_{d|n}g(n/d)$ – 2017-01-25
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    This is not what applying the first direction yields. Or is it? – 2017-01-25
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    What do you mean ? – 2017-01-25
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    If $f(n)=\sum_{d|n} g(d), \quad$then $\quad g(n)=\sum_{d|n} \mu(d)f(n/d)$, applying this to $r(n)=\sum_{d|n}\mu(n/d)s(d)$ yields $\mu(n/n)s(n)=\sum_{d|n}\mu(d)r(n/d)$ – 2017-01-25
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    No. it yields $s(n) = \sum_{d | n} r(d) = \sum_{d | n}r(n/d)$ – 2017-01-25
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    I'm sorry... but how? Is my substitution wrong? – 2017-01-25
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    Come on... Yes it is wrong. Do you realize $\sum_{d |n} \mu(d) s(d)$ is not the same as $\sum_{d |n} \mu(n/d) s(d)$ ? – 2017-01-25
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    I see it now, thanks! – 2017-01-25

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As you know $\sum_{d | n} \mu(d)= 1_{n = 1}$ so that $$\sum_{d | n} \mu(d) f(\frac{n}{d}) = \sum_{d | n} \mu(d) \sum_{m | \frac{n}{d}} g(m)= \sum_{d | n}\sum_{m | \frac{n}{d}} \mu(d) g(m)$$ $$= \sum_{m | n } \sum_{d | \frac{n}{m}}g(m) \mu(d)=\sum_{m | n } g(m)1_{\frac{n}{m}=1} = g(n) $$