Definition of isomorphism of graded rings After searching through some literature I got a bit confused what one has to check to conclude that two graded rings are isomorphic (as graded rings).
Suppose that $R$ and $S$ are graded rings, then a homomorphism of graded rings is a ring homomorphism 
$$f:R\rightarrow S$$ such that $f(R_{n})\subset S_{n}$ for all $n$.  
Question: Suppose that we have such a homomorphism of graded rings between $R$ and $S$, then to conclude that $R$ and $S$ are isomorphic as graded rings, is it then enough to show that the map $f$ from above is a ring isomorphism? Or do we also have to show that $f(R_{n})=S_{n}$ for all $n$? 
 A: An isomorphism of graded rings is a morphism $f:R \rightarrow S$ of graded rings, such that there is an inverse morphism $g:S \rightarrow R$ of graded rings. So in particular $f$ and $g$ are isomorphism of rings.
Conversely assume we have a graded morphism of rings $f:R 
\rightarrow 
S$, which is an isomorphism of rings ie. there is an inverse morphism of rings $g:R\rightarrow S$. Now the only thing to check is that $g$ is indeed a morphism of graded rings meaning $g(S_n)\subseteq R_n$. Now since $f$ is graded with inverse map $g$ we find that $g(S_n) = f^{-1}(S_n) = R_n$. This shows that $g$ is an morphism of graded rings, hence $f$ is an isomorphism of graded rings. With other words, the forgetful functor $\mathsf{grRing} \rightarrow \mathsf{Ring}$ is conservative (reflects isomorphisms).
A: An isomorphism of graded rings (or anything else really) is a homomorphism with an inverse such that the inverse is also a homomorphism.
You can prove that a ring isomorphism that is an homomorphism of graded rings is an isomorphism of graded rings. To see this think about where $f^{-1}$ maps $S_n$, or alternatively think about what it would mean for $f(R_n)$ to not equal $S_n$.
