What does it mean to induce a functor?

Question

I have a text ("Sheaf Theory Through Examples") that just drops some weird notation and the concept of inducing a functor here:

"In other words, composing with $t$ induces a functor $\textbf{C} \cong Fun(1, \textbf{C}) \to Fun(\textbf{J}, \textbf{C})$, which is commonly denoted $\Delta_t: C \to Fun(\textbf{J}, \textbf{C}) = \textbf{C}^\textbf{J}$."

I don't really know what is meant with Fun(a, b) here because it wasn't defined yet. And composing seems to 'induce a functor' but it's not clear what is meant by it.

Answer 1

$Fun(1,C)$ denotes functors from $1$ to $C$. $1$ has one object and only identity morphism.

Similarly with $J$ but I'm guessing $J$ denotes some sort of diagram category. Like $3$ objects with non-identity morphisms in a vee shape. You can draw dots decorated with objects of $C$. One such choice is decorating them all the same in analogy with the constant function if $J$ and $C$ were just sets.

You have $t$ as functor from $J$ to $1$. It is some object of $Fun(J,1)$. You can compose to reach $Fun(J,C)$ starting from $Fun(1,C)$. That is what is being denoted as $\Delta_t$.

See what it does on an object $x$ of $C$. That is also regarded as the functor $1 \to C$ given by sending $1 \to x$ and $id_1 \to id_x$. Composition with $t$ gives the functor which sends all objects of $J$ to $1$ and then to $x$. The morphisms of $J$ collapse too. So you can assure yourself that what you get is an object of $Fun(J,C)$.

See what it does on a morphism $f$ in $C$. That is a natural transformation of two $1 \to C$ functors. Then you want that composition $\Delta_t (f)$ to be a natural transformation of the corresponding $J \to C$ functors. Like checking a square that comes from the morphism $g$ in $J$ you have a square with $x$, and $y$ on the corners and $f$ on the side edges.

Then after seeing what it does, you can check that it satisfies the laws for identities and compositions.

This entire thing is like the constant functions between sets $X \to Y$ that are determined by elements $y$ of a set $Y$ by doing $X \to \text{Singleton} \to Y$.

Draw pictures with $J$ being something tiny like said with the vee shape or similar.

Answer 2

This is not a detailed answer but rather a long comment:

If $C$ and $D$ are categories, $Fun(C,D)$ typically denotes the category of functors between $C$ and $D$. Its objects are all such functors, while its morphisms are the natural transformations between them. Notice that a functor is not only of objects and morphisms, but it satisfies two extra rules: Images of the identities go to identities and images of compositions are composition of images. I'm sure $t$ assigns objects and morphisms of $\mathbf{C}$ to objects and morphisms in $\mathbf{J}$, resp. Saying that it induces a functor means that $t$ satisfies those extra rule I have just mentioned, i.e. preserves identities and composition.