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Is it true that the centre of a category is just the set of identity morphisms? I'm reading through notes and I'm not able to able to see the reason for defining the category of functors and calling the centre the natural transformations from the identity to identity

2 Answers 2

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No, you've misunderstood - the center of a category $C$ is not the set $\{\mathrm{Id}_x:x\in C\}$.

From the nLab article (link):

Specifically, the center of a category $C$ is defined to be the commutative monoid $[C,C](\mathrm{Id}_C,\mathrm{Id}_C)$ of endo-natural-transformations of the identity functor of $C$.

Side note: If $x$ and $y$ are objects of a category $C$, then $C(x,y)$ is a common notation for the collection of morphisms in $C$ from $x$ to $y$. Moreover, $[C,D]$ is a common notation for the category of functors from $C$ to $D$, which is defined as having as its objects the functors $F:C\to D$, and as its morphisms from an object $F:C\to D$ to an object $G:C\to D$, the natural transformations $\alpha:F\Rightarrow G$.

Thus, $[C,C]$ is the category of functors from $C$ to itself, and the identity functor $\mathrm{Id}_C$ is an object of $[C,C]$, and the notation $[C,C](\mathrm{Id}_C,\mathrm{Id}_C)$ means the collection of morphisms in $[C,C]$ from the object $\mathrm{Id}_C$ to itself, i.e., all the natural transformations from the functor $\mathrm{Id}_C$ to itself.

Also from the nLab article:

It is straightforward to check that this reduces to the usual definition if $C = \mathbf{B}(A,\times)$ is the delooping of a monoid.

That is, if we start with a monoid $A$, and we define a category $C$ to have a single object $\ast$ with morphisms $C(\ast,\ast)=A$, then $[C,C]$ is equivalent to the collection of monoid homomorphisms $A\to A$, with $\mathrm{Id}_C$ being the identity homomorphism $A\to A$, and $[C,C](\mathrm{Id}_C,\mathrm{Id}_C)$ amounts to the center of the monoid $A$.

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No, it's not true.

Example: An illustrative example is the category $R\text{-Mod}$ of modules over a commutative ring $R$. Spelling out the definition of center in this case, it consists of families of endomorphisms $\varphi_M: M\to M$ indexed over $R$-modules $M$ that are natural with respect to any $R$-linear homomorphism $M\to N$. Now, by definition, $R$-linear homomorphisms commute with multiplications by scalars from $R$, so for any $r\in R$ the family $\varphi(r)$ defined by $\varphi(r)_M: M\to M, m\mapsto rm$ provides an element in the center of $R\text{-Mod}$, which for $r\neq 1$ is different from the family of identity morphisms.

Exercise: Show that this even defines an isomorphism between $R$ and the center of $R\text{-Mod}$.

Exercise: Think about what happens for a not necessarily commutative ring.