Induction and Coinduction are isomorphic when $[G:H] < \infty$ If $[G:H]<\infty $ then $Ind_H^G M \cong Coind_H^G M$.
There is an $H$-map $\phi_0 : M \rightarrow Hom_H(\mathbb{Z}[G],M)$ given by
$ \phi_0(m)(g) =  gm $ if $g\in H  ,0   g\notin H$.
Now this will extend to a $G$-map $\phi: \mathbb{Z}[G] \otimes_{\mathbb{Z}[H]} M \rightarrow Hom_{\mathbb{Z}}(\mathbb{Z}[G],M)$, and we will have that $\phi^{-1}(f)=\sum_{g \in G/H}g\otimes  f(g^{-1})$.
Im am trying to see that these maps are actually inverse one another but I am failing so when help is welcomed. 
Another thing is that when we define the inverse the sum is running trough the equivalence classes how do we know that this is independent on the choice of representative, is it because $f$ is an $H$-homomorphism so we will have on side of the tensor product $H$ and on the other $H^{-1}$ and they will cancel out?
 A: First, of all the map $$\phi : \mathbb{Z}[G] \otimes_{\mathbb{Z}[H]} M \to \text{Hom}_H(\mathbb{Z}[G], M)$$
induced by $\phi_0$ is given by 
$$ \phi(g \otimes m)(g')  = (g \phi_0(m))(g') = \phi_0(m)(g' g)$$
for $g,g' \in G$ and $m \in M$
Let us denote the hopefully inverse map by
$$ \psi : \text{Hom}_H(\mathbb{Z}[G], M) \to  \mathbb{Z}[G] \otimes_{\mathbb{Z}[H]} M, f \mapsto \sum_{g \in G/H} g \otimes f(g^{-1}).$$
As you rightfully noted, you first have to check that this is actually well-defined, that is, independent of the choice of coset representatives.
Indeed for any $g \in G$ and any $h \in H$ we have 
$$(gh) \otimes f((gh)^{-1}) = g h \otimes h^{-1} f(g^{-1}) = g \otimes f(g^{-1})$$
since $f$ is an $\mathbb{Z}[H]$-module homomorphism and because we are taking the tensor product over $\mathbb{Z}[H]$.
So we can take any transversal $T \subseteq G$ of$G/H$ and obtain $\psi(f) = \sum_{g \in T} g \otimes f(g^{-1})$.
Now we have to calculate the compositions $\psi \circ \phi$ and $\phi \circ \psi$.
We have 
$$ (\psi \circ \phi)(g \otimes m) = \sum_{t \in T} t \otimes \phi_0(m)(t^{-1}g) $$
Note that $\phi_0(m)(t^{-1}g) = 0$ unless $g = t h$ for some $h \in H$. Since cosets are disjoint, it follows that only one summand is nonzero and with $g =  th$ we get 
$$ (\psi \circ \phi)(g \otimes m) = t \otimes \phi_0(m)(t^{-1}g)
                                  = t \otimes h m = th \otimes m = g \otimes m $$ 
Conversely, 
$$ (\phi \circ \psi)(f)(g) = \sum_{t \in T} \phi(t \otimes f(t^{-1}))(g) = \sum_{t \in T} \phi_0(f(t^{-1}))(gt) $$ 
and again all summands vanish except for the one with $gt \in H$ (or $g^{-1} \in tH$). With $gt = h$ for some $h \in H$ we then obtain 
$$ (\phi \circ \psi)(f)(g) = \phi_0(f(t^{-1}))(h) = h f(t^{-1}) = f(ht^{-1}) = f(g)$$
and thus both compositions yield the respective identity morphisms.
