Question about finding the nth root in a modulus. Some notation:
$\mathbb Z_n$ denotes the set of integers up to n i.e.  $\mathbb Z_n=\{i│i ∈ \mathbb Z,0≤i < n  \}$. 
$\mathbb Z_n^*$ denotes invertible elements of $\mathbb Z_n$   i.e.  $\mathbb Z_n^*=\{i|i ∈ \mathbb Z,0 ≤ i < n,gcd⁡(i,n)=1  \}$. 
Suppose:
$m$ is a member of $\mathbb Z_n$,
$n$ is a composite of two large primes $p$ and $q$.
$g$ is a member of $\mathbb Z_{n^2}^*$, where $g$'s order in modulo $n^2$ is divisible by $n$.
$r$ is a member of $\mathbb Z_n^*$
then, it has been proven
 that the function $P(m,r) = g^m \cdot r^n \bmod n^2$ is a bijection $(\mathbb Z_n  \times \mathbb Z_n^* → \mathbb Z_{n^2}^*)$ 
my question is, given that one knows the factorization of $n=p\cdot q$, $g,m,n$ can one efficiently find $r$?
 A: I assume that the question is: Given $p,q,n,g,m$ and an element $x=P(m,r)$, one wants to have a method for finding $r$.
First I need to redefine the mappings so that they make sense to me. Please bear with me, this is just me tuning up. I will get to the answer. But I also need the following extra assumptions: $$\gcd(p,q-1)=\gcd(p,q)=\gcd(p-1,q)=1.$$ These assumptions are not overly restrictive, and most likely e.g. won't slow down key generation noticeably.
Much of the arithmetic takes place in the group $G=\mathbb{Z}_{n^2}^*$ that has size $pq(p-1)(q-1)$. This has a subgroup 
$$U=(1+n\mathbb{Z}_{n^2})=\{\overline{a}\in\mathbb{Z}_{n^2}\mid a\equiv1\pmod{n}\}$$ of order $n$.
There is an injective homomorphism $f:\mathbb{Z}_n^*\rightarrow G$ defined as follows. Let $r$ be an integer coprime to $n$. If $r\equiv r'\pmod n$, then a well-known calculation shows that $r^p\equiv r'^p\pmod{p^2}$, and hence $r^{pq}=r^n\equiv r'^n\pmod{p^2}$. By changing the roles of $p$ and $q$ in the preceding argument, we similarly see that $r^n\equiv r'^n\pmod{q^2}$. By Chinese Remainder Theorem we deduce that $r^n\equiv r'^n\pmod{n^2}$. The mapping $f:\overline{r}\mapsto\overline{r^n}$ is thus well-defined, and it is obviously a homomorphism of groups. Let us write $H=Im(f)$. 
The homomorphism $f$ is actually injective. This follows (in the spirit of Lemma 13 from the seminar notes linked to by OP) from the facts that $H=G^n$ and the $U$ is the kernel of the $n$th power mapping $G\rightarrow G$ (see the link). A calculation of orders shows that we have a direct product decomposition
$$
G=U\times H.
$$
If $ord_G(g)$ is a multiple of $n$, then the elements $g^i, i=1,2,\ldots,n-1$ all have orders that are divisible by at least one of $p$ or $q$, and thus won't belong $H$, so the powers $g^i=\{0,1,\ldots,n-1\}$ form a complete set of representatives of cosets of $H$. Thus (Theorem 16 of the linked seminar notes) the mapping
$$
P:\{0,1,2,\ldots,n-1\}\times \mathbb{Z}_n^*\rightarrow \mathbb{Z}_{n^2}^*, (m,r)\mapsto g^m f(r)
$$
is a bijection.
Now we can reformulate the question to read: Given $x\in G$ that is known to be in the coset $g^m H$, how can we find $r\in \mathbb{Z}_n^*$ such that $x=g^mf(r)$, if we have the (trapdoor) information that we know $p$ and $q$. By CRT it suffices to compute $r$ modulo $p$ and modulo $q$. We first compute $f(r)=g^{-m}x$. Here $r^p\equiv r\pmod p$ by Little Fermat, so $f(r)\equiv r^n\equiv r^q\pmod p$. Let $i_1$ be the modular inverse of $q$ modulo $p-1$ (I need the extra assumption $gcd(p-1,q)=1$ here!), so $i_1q= 1+k(p-1)$ for some integer $k$. Therefore
$$
f(r)^{i_1}\equiv (r^q)^{i_1}=r^{1+k(p-1)}=r\cdot r^{k(p-1)}\equiv r\pmod{p}.
$$
Similarly, if $i_2$ is the modular inverse of $p$ modulo $q-1$, we get that
$$
f(r)^{i_2}\equiv r\pmod{q}.
$$
Thus in order to compute $r$ modulo $n$, we need to raise $f(r)$ to the powers $i_1$ and $i_2$, compute their respective residues modulo $p$ and $q$, and apply the CRT to the results.
