In category theory, a presheaf on a category $C$ is a functor $F: C^{op}\longrightarrow Set$. If $C$ is the poset of open sets in a topological space, interpreted as a category, then one recovers the usual notion of presheaf on a topological space.
When there is a topological space $\mathscr{T}$, there is a corresponding Ring for each open set.
I know that we can think of a continuous function as an example for presheaf like above picture.
There is a continuous function defined in the open interval $(-2,2)$, and even if this function is reduced to a smaller $(-1,1)$, it becomes a continuous function. And a collection of continuous functions becomes a Ring.
We can express this relationship as follows:
$(-2,2) \longrightarrow \mathscr{F}(-2,2)$
$\downarrow$
$(-1,1) \longrightarrow \mathscr{F}(-1,1)$
(In this diagram $\downarrow$ means restriction.)
However, when I think of presheaf as a continuous function like this, I wonder how I should think about it from the perspective of a category theory.
Best Answer
I am guessing you are referring to the following, I split it up into two parts.
Topology:
Let $X$ be a topological space, for example the space $(-2,2)\subset \mathbb{R}$ as above. For an open subset $U\subseteq X$, one defines the set $\mathrm{Cont}(U,\mathbb{R})$ of continuous maps $f\colon U \to \mathbb{R}$. As a general principle: The restriction of a continuous function is continuous. To an inclusion of opens $i\colon U\to V$ we may assign the function of sets $$ i^*=\mathrm{res}\colon \mathrm{Cont}(V,\mathbb{R})\to \mathrm{Cont}(U,\mathbb{R}), $$ which precomposes with $i$. Note also the reversal of the order of $U$ and $V$. This assignment from inclusions to restrictions is compatible with composition.
Category theory:
We define a category $\mathbf{Ouv}(X)$ coming from the poset of open subsets of $X$, ordered by inclusion. We now define a functor $\mathcal{F}\colon \mathbf{Ouv}(X)^{\mathrm{op}}\to \mathbf{Set}$ as follows:
On objects, it should send an open subset $U$ to the set $\mathcal{F}(U) = \mathrm{Cont}(U,\mathbb{R})$. Now onto morphisms: in $\mathbf{Ouv}(X)^{\mathrm{op}}$ there is a morphism $V\to U$ if and only if there is an inclusion of open sets $U\subseteq V$. In this case we assign to it the morphism $$ \mathrm{res}\colon \mathrm{Cont}(V,\mathbb{R})\to \mathrm{Cont}(U,\mathbb{R}) $$ of sets. It is now easy to check that this is a functor, by the compatibility of the restriction maps we noticed above.
To your questions in the comments:
What I have described above is a sheaf of sets, often simply called a sheaf. What you seem to be referencing is a sheaf of rings, a functor $\mathcal{G}\colon \mathbf{Ouv}(X)^{\mathrm{op}}\to \mathbf{Ring}$. We may upgrade our functor $\mathcal{F}$ to such a sheaf as follows. Noting that the objects $\mathcal{F}(U) = \mathrm{Cont}(U,\mathbb{R})$ are already rings in the obvious way, and that the restriction maps are already maps of rings, we may define $\mathcal{F'}\colon\mathbf{Ouv}(X)^{\mathrm{op}}\to \mathbf{Ring}$ like this on objects, and the same way as $\mathcal{F}$ on morphisms. Now with $i\colon (-1,1)\to (-2,2)$ and $$ f\colon (-2,2)\to \mathbb{R}, \quad x \mapsto x^3-4x+5 $$ we have $\mathcal{F}'(i^{\mathrm{op}})(f) = f\mid_{(-1,1)}$.
There was a typo which I have fixed.