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Given a category $\mathcal{C}$, the functorial morphisms $\phi:\rm{id}_{\mathcal{C}}\rightarrow \rm{id}_{\mathcal{C}}$ define a monoid via $(\phi\circ \psi)_X := \phi_X\circ\psi_X$ for $X\in \rm{Ob}(\mathcal{C})$, where the unit is $\rm{id}_{\rm{id}_\mathcal{C}}:\rm{id}_{\mathcal{C}}\rightarrow \rm{id}_{\mathcal{C}}$, $ (\rm{id}_{\rm{id}_\mathcal{C}})_X := \rm{id}_X$.

The goal is to compute this monoid for the categories $\rm{Set}$, $\rm{Ring}$ and $R-\rm{Mod}$ for a given commutative ring $R$.

How do I approach this? For $\rm{Set}$, let $\phi$ be such a functorial morphism, $X,Y$ be any two sets. Then there is a commutative diagram such that $f\circ \phi_X = \phi_Y \circ f$ for every map $f:X\rightarrow Y$. I want to conclude some necessary properties of $\phi_X$ from this, but I don't know how yet.

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If you take $X = *$, the one element set, then the commutative diagram of $\phi$ and $f$

enter image description here

becomes

enter image description here

since $\phi_X = \text{id}_X$. And thus since functions $* \rightarrow Y$ are in bijection with elements of $Y$ we can conclude that $\phi_Y$ has to fix every point in $Y$. Thus $\phi$ has to be the identity natural transformation. Essentially what we used here is that morphisms $* \rightarrow Y$ are in bijection with elements of $Y$.

You can also use the Yoneda lemma for this, the identity functor on $\text{Set}$ is naturally isomorphic to $\text{Hom}(*,-)$ and by yoneda lemma $\text{Nat}(\text{Hom}(*,-),\text{Hom}(*,-)) \approx \text{Hom}(*,*) \approx *$

In the case of the category $R-\text{Mod}$, what module $I$ has the property that module homomorphisms $I \rightarrow M$ are in bijection with elements of $M$? After you figure that out, $I$ should play a similar role in your argument as that of $*$ in $\text{Set}$. What do module homomorphisms $I \rightarrow I$ look like?

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  • $\begingroup$ If I consider $R$ as an $R$\,-\,module and consider $R$ - module homomorphisms $f:R\rightarrow M$ into any $R$ - module $M$, we have $f(r) = r f(1)$ and hence $\rm{Hom}(R,M)$ corresponds bijectively to points in $M$. Thus, $\rm{Hom}(R,R) \eqsim R$. By $f\circ \phi_R = \phi_M \circ f$ I get that $\phi_M$ must have the form $\phi_M(x) = a\cdot x$ for some $a\in R$. The element $a = \phi_R(1)$ is independent of the choice of $M$ and only depends on the choice of $\phi_R$. Thus, the functorial morphisms of the identity correspond to points in $R$. Correct? $\endgroup$ – Teddyboer Apr 6 at 12:58
  • $\begingroup$ Yes that is correct! You just need to show that the correspondance is a monoid homomorphism, regarding $R$ as a monoid defined by its multiplication. $\endgroup$ – Noel Lundström Apr 6 at 13:53
  • $\begingroup$ Okay and I think I see why the latter is true. Thank you. $\endgroup$ – Teddyboer Apr 6 at 14:07
  • $\begingroup$ Well it's easier to show that the function $R \rightarrow \text{Nat}(id,id)$ is a homomorphism rather than than the map $\text{Nat}(id,id) \rightarrow R$ $\endgroup$ – Noel Lundström Apr 6 at 15:21
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Noel Lundström has already given the answer when the identity functor is representable, as is the case in $\newcommand\Set{\mathbf{Set}}\newcommand\Mod{\mathbf{Mod}}\Set$ and $R\text{-}\Mod$. The question still remains of what to do in $\newcommand\Ring{\mathbf{Ring}}\Ring$, since the identity is not representable anymore.

The solution here is partly given by the same idea. Forget down from $\Ring$ to $\Set$. Call the forgetful functor $U$. $U$ is representable, we have $U\cong \Ring(\Bbb{Z}[x],-)$.

Moreover, $U$ is faithful, since a ring homomorphism is determined by what it does to elements of the ring. Suppose then that $\phi :1_\Ring\to 1_\Ring$ is an endomorphism of the identity functor, then applying $U$ we have $U\phi : U\to U$ is an endomorphism of $U$, and moreover $U(\phi\psi) = (U\phi)(U\psi)$, since $U$ is a functor, and finally if $U\phi = U\psi$, then $\phi=\psi$, since $U$ is faithful. Therefore $U$ embeds the monoid of endomorphisms of the identity functor into the monoid of endomorphisms of $U$.

By the Yoneda lemma, the endomorphisms of $U$ are given by $\Ring(\Bbb{Z}[x],\Bbb{Z}[x])$, which can be identified with the monoid of one variable polynomials under composition of polynomials, i.e. $p*q = p(q(x))$. Note that $x$ is the identity.

Now we're just left with the question of which endomorphisms come from endomorphisms of the identity functor of rings.

Well, the natural transformation corresponding to the polynomial $p(x)$ is the map of sets $r\mapsto p(r)$ for $r\in R$ an element of any ring $R$. When does this define a ring homomorphism?

Well, let's consider $\Bbb{Z}$. $\Bbb{Z}$ has no nontrivial ring homomorphisms from it to itself, since $1$ generates $\Bbb{Z}$ and $1$ must be fixed by any ring homomorphism. Thus $p(n)=n$ for all integers $n$. But then $p(x)-x$ has infinitely many zeros, so $p(x)=x$. Thus the only endomorphism of $1_\Ring$ is the identity.

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