$$
\newcommand{cat}[1]{\operatorname{#1}}
$$
I am working through the following excersice found in Emily Riehl's Category Theory in context,
Exercise 2.2.vi. Do there exist any non-identity natural endomorphisms of the category
of spaces? That is, does there exist any family of continuous maps $X \longrightarrow X$, defined for all
spaces $X$ and not all of which are identities, that are natural in all maps in the category $\operatorname{Top}$?
and want to check if my reasoning is correct.
Here's what I've thought: if such a natural transformation $\eta : id_{\cat{Top}} \Rightarrow id_{\cat{Top}}$ would exist, with at least one non trivial arrow, this would give rise to a natural endomorphism of the forgetful functor $U: \cat{Top} \rightarrow \cat{Set}$ by composing with such functor, whose arrows are not all trivial. In particular, we would have that $\cat{Nat}(U,U)$ has more than one element, because there also exists the trivial natural transformation. However, the forgetful functor $U$ is representable by the singleton space $*$ and so
$$
\cat{Nat}(U,U) \simeq \cat{Nat}(\cat{Top}(*,-), \cat{Top}(*,-)) \stackrel{Yoneda}\simeq \cat{Hom}_{\cat{Top}}(*,*)
$$
which is absurd, because it would imply
$$
1 < |\cat{Nat}(U,U)| = |\cat{Hom}_{\cat{Top}}(*,*)| = 1
$$
Have I interpreted the question correctly? I'm having a hard time processing some statements.
Best Answer
Yes, this is correct. More directly, you can just consider any space $X$, any $x\in X$, and the map $*\to X$ which sends the point to $x$. Naturality of $\eta$ with respect to this map says exactly that $\eta_X(x)=x$. (This argument is really exactly the same as yours, though, when you unwind the proof of Yoneda's lemma.)