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Stream: learning: questions

Topic: ✔ Does evaluation of a functor at object define a functor?

Ignat Insarov (Apr 05 2024 at 05:20):

Say we have an object $X$ in $\mathbb{C}$.

• For any functor $F: \mathbb{C} → \mathbb{D}$, we can compute $FX$, thereby sending objects of $\mathbb{D}^{\mathbb{C}}$ to objects of $\mathbb{D}$.
• For any natural transformation $τ: F → G$, we can compute its component at $X$ — $τ_X: FX → GX$.
• Since identity and composition of natural transformations is defined component-wise, we have $υ = τ ∘ σ ⇒ υ_X = τ_X ∘ σ_X$ and $id_X$ is the identity of $X$.

So, for any object $X$ in $\mathbb{C}$ we can have a functor $(·)X: \mathbb{D}^{\mathbb{C}} → \mathbb{D}$ defined as $F ↦ FX, τ ↦ τ_X$.

Is this right?

Ralph Sarkis (Apr 05 2024 at 05:43):

That is correct, I have seen this functor called the evaluation functor at $X$.
The evaluation functor is what you get when you consider all $X$ at once, it is of type $\mathsf{ev}: \mathbb{C} \times \mathbb{D}^{\mathbb{C}} \rightarrow \mathbb{D}$.

Notification Bot (Apr 05 2024 at 09:08):

Ignat Insarov has marked this topic as resolved.

Mike Shulman (Apr 05 2024 at 15:01):

Ralph Sarkis said:

The evaluation functor is what you get when you consider all $X$ at once, it is of type $\mathsf{ev}: \mathbb{C} \times \mathbb{D}^{\mathbb{C}} \rightarrow \mathbb{D}$.

Or $\mathsf{ev}: \mathbb{D}^{\mathbb{C}} \times \mathbb{C} \rightarrow \mathbb{D}$, since we usually write $F(X)$ rather than $(X)F$.