Showing 2 notions are equivalent When an exercise asks you to show 2 notions are equivalent, what is it really asking you to do?
For example in Hartshorne chapter 2 exercise 2.7 he asks us to show that given a scheme $(X, O_X)$ and a field $K$ and for $x \in X$ denoting by $k(x)$ the field $O_x/m_x$ where $O_x$ is the stalk of the sheaf $O_X$ at $x$, that to give a morphism of $Spec K$ to $X$ is equivalent to giving a point $x \in X$ and an inclusion $k(x) \rightarrow K$.
The reason I ask is that I've seen solutions to questions like these which (in the context of my  example for clarity's sake) given a morphism will show that you can create a point and the inclusion map, and conversely given the point and map show you can create the morphism and just do that.
However I remember a lecturer giving me a question like this and in his solution he also shows that going from one notion to the other then back again results in the original thing (and does this both ways).
This extra bit of work seems to make sense if we are trying to establish perhaps a bijection between the sets of the 2 notions in the question but I am unsure if that is what these questions are asking.
 A: Let's phrase this more precisely: the general form of such questions is 'given a mathematical object $X$, specifying a structure $\sigma_1$ on $X$ of one kind is equivalent to specifying a structure $\sigma_2$ on $X$ of another kind'. Let $S_1$ be the set of structures on $X$ of the first kind and let $S_2$ be the set of structures on $X$ of the second kind.
If specifying an $S_1$-structure on $X$ is to be equivalent to specifying an $S_2$-structure on $X$, then:


*

*Each $\sigma_1 \in S_1$ can be obtained by specifying some $\sigma_2 \in S_2$ and applying some procedure to $\sigma_2$, which means that you have an injective function $S_1 \to S_2$ (which sends $\sigma_1$ to some choice of $\sigma_2$ which determines it); and

*Each $\sigma_2 \in S_2$ can be obtained by specifying some $\sigma_1 \in S_1$ and applying some procedure to $\sigma_1$, which means that you have an injective function $S_2 \to S_1$.


I see no reason in general why these functions should be mutually inverse (thereby establishing a bijection), but if they are, then that's a bonus! But if they're not mutually inverse, then they'd better be injective. In any case, the Cantor–Schröder–Bernstein theorem then implies that a bijection $S_1 \to S_2$ exists, even if you haven't explicitly defined it.
In more practical terms, this amounts to saying that each $S_1$-structure gives rise to at most one $S_2$-structure, and vice versa, but the procedures for going in each direction might be different (and, in particular, might not be mutually inverse).
A: If you want to phrase that in the language of categories. Consider the category $\mathsf{Sch}$ of schemes, the category $\mathsf{Set}$ of sets, and the category $\mathsf{Fld}$ of fields. You have two functors* $\mathsf{Sch} \to \mathsf{Set}$, given by:
$$\begin{align}
F : X & \mapsto \operatorname{Hom}_{\mathsf{Sch}}(\operatorname{Spec} K, X) \\
G : X & \mapsto \{ (x, i) \mid x \in X, i \in \operatorname{Hom}_{\mathsf{Fld}}(k(x), K)
\end{align}$$
what Hartshorne is asking you to do is to prove that there two functors are naturally isomorphic, i.e. finding an isomorphism (a bijection) $\eta_X : F(X) \to G(X)$ which commutes with all natural maps. Pretty much any problem of the form you are describing can be rephrased this way.
(What your lecturer did was describing $\eta : F \to G$ but also the inverse natural isomorphism $\theta : G \to F$, and proving that $\eta \circ \theta = \operatorname{id}$ and $\theta \circ \eta = \operatorname{id}$. Of course strictly speaking you don't need to do that, but what better way to prove that something is an isomorphism than to explicitly describe its inverse isomorphism?)

* Strictly speaking I'm only describing what the functors do on objects. You must also describe what it does on morphisms, i.e. if $f : X \to Y$ is a morphism of schemes, what's the induced function $F(X) \to F(Y)$ (resp. $G(X) \to G(Y)$)? I leave that as an exercise.
