Finite-group induction is coinduction, hence biadjoint to restriction

${\mathrm{\Phi }}_M$ is a well-defined $kG$-linear map. For fixed $g,m$ the formula defines a function on $G$: if $xg\in H$ then $(xg)m\in M$, otherwise $0$. It is $k$-linear in $x$ by definition, and it is $kH$-linear: for $h\in H$ and $x\in G$, ${\mathrm{\Phi }}_M(g\otimes m)(hx)=((hx)g)m$ if $hxg\in H$ and $0$ otherwise; since $hxg=(hx)g$ and $hxg\in H⟺xg\in H$ (as $h,h^{-1}\in H$), this equals $h((xg)m)=h{\mathrm{\Phi }}_M(g\otimes m)(x)$ in the first case and $0$ in the second. Hence ${\mathrm{\Phi }}_M(g\otimes m)\in {\mathrm{Coind}}_H^{G}M$. The map is $k$-linear and in the second tensor variable balanced: for $h\in H$, the functions ${\mathrm{\Phi }}_M(gh\otimes m)$ and ${\mathrm{\Phi }}_M(g\otimes hm)$ both send $x$ to $((xg)h)m=(xg)(hm)$ when $xg\in H$ and to $0$ otherwise; so ${\mathrm{\Phi }}_M$ is well defined on the tensor product. Finally it is $kG$-linear: ${\mathrm{\Phi }}_M(g_{0}g\otimes m)(x)=((x{g}_0)g)m$ if $x{g}_{0}g\in H$ and $0$ otherwise, and $(g_0{\mathrm{\Phi }}_M(g\otimes m))(x)={\mathrm{\Phi }}_M(g\otimes m)(x{g}_0)=((x{g}_0)g)m$ if $x{g}_{0}g\in H$ and $0$ otherwise, the two conditions being the same.

The inverse on coinduced maps. For $\phi \in {\mathrm{Coind}}_H^{G}M$, define ${\mathrm{\Psi }}_M(\phi )={\sum }_{t\in T}t\otimes \phi (t^{-1})$. This is a finite sum and a $k$-linear map. Inverting the left cosets $tH$ shows that the right cosets $H{t}^{-1}$ partition $G$. If a representative is replaced by $t^{\mathrm{\prime }}=th$ with $h\in H$, then $t^{\mathrm{\prime }}\otimes \phi (t^{\mathrm{\prime }-1})=th\otimes \phi (h^{-1}{t}^{-1})=th\otimes h^{-1}\phi (t^{-1})=t\otimes \phi (t^{-1})$, by $kH$-linearity and the balancing relation. Thus the inverse formula is independent of the finite transversal.

${\mathrm{\Phi }}_M{\mathrm{\Psi }}_M=\mathrm{i}\mathrm{d}$. For $x\in G$, there is exactly one $t_0\in T$ for which $x\in H{t}_0^{-1}$, equivalently $x{t}_0\in H$. Write $x=h{t}_0^{-1}$. Then ${\mathrm{\Phi }}_M({\mathrm{\Psi }}_M(\phi ))(x)={\sum }_{t\in T,xt\in H}(xt)\phi (t^{-1})=h\phi (t_0^{-1})=\phi (x)$, using the right-coset partition from step 1.2 and $kH$-linearity. Equality on the group basis gives equality on $kG$.

${\mathrm{\Psi }}_M{\mathrm{\Phi }}_M=\mathrm{i}\mathrm{d}$. For $g\in G$ and $m\in M$, write uniquely $g=t_{0}h$ with $t_0\in T$ and $h\in H$. Then ${\mathrm{\Psi }}_M({\mathrm{\Phi }}_M(g\otimes m))={\sum }_{t\in T,t^{-1}g\in H}t\otimes (t^{-1}g)m=t_0\otimes hm=g\otimes m$, by the left-coset partition and tensor balancing. Since such elementary tensors span the induced module, this proves the second inverse identity.

Naturality and consequences. If $f:M\to M^{\mathrm{\prime }}$ is a $kH$-map, then for all $g,m,x$ one has ${\mathrm{\Phi }}_{M^{\mathrm{\prime }}}(g\otimes f(m))(x)=f({\mathrm{\Phi }}_M(g\otimes m)(x))$, because the formula is evaluated by multiplication in $M$ or $M^{\mathrm{\prime }}$ and $f$ is linear; hence $\mathrm{\Phi }$ is a natural transformation, and by steps 2.1 and 2.2 each ${\mathrm{\Phi }}_M$ is an isomorphism with natural inverse ${\mathrm{\Psi }}_M$. Therefore ${\mathrm{Ind}}_H^G\cong {\mathrm{Coind}}_H^G$ as functors, and [F2] yields that induction is both a left and a right adjoint of restriction: the left adjunction is its induction adjunction, and the right adjunction is transported across this natural isomorphism from the coinduction adjunction. Induction $kG{\otimes }_{kH}-$ is right exact, and coinduction ${\mathrm{Hom}}_{kH}(kG,-)$ is left exact; a natural isomorphism between the two functors therefore makes induction exact, in agreement with the freeness of $kG$ over $kH$ recorded in [F3]. No hypothesis on $k$, on characteristic, or on splitting is used, and the only selection made is the finite transversal $T$ of least representatives.