# Functorial morphisms on the identity functor

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.

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

becomes

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?

• 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? – Teddyboer Apr 6 at 12:58
• 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. – Noel Lundström Apr 6 at 13:53
• Okay and I think I see why the latter is true. Thank you. – Teddyboer Apr 6 at 14:07
• Well it's easier to show that the function $R \rightarrow \text{Nat}(id,id)$ is a homomorphism rather than than the map $\text{Nat}(id,id) \rightarrow R$ – Noel Lundström Apr 6 at 15:21


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.