I've been trying to understand an analogue of Russell's paradox for categories (Please refer to this link for more details and preliminaries about my question: Russell Pardox for naive category theory)
I don't understand the first lemma, specifically how does the presence of M and N in $C_1$ results in any arbitrary category in $C_1$ being isomorphic to some category in $C_2$? I feel like something was skipped in the sketch which is needed to understand it. So, it will be great if someone could shine light on this missing thing.
(The relevant part of the text is also copied here.)
Just for reference, let me give some more details of my Russell paradox for naive category theory.
Setting:
We are working in a naive category theory setting, i.e., category theory where the foundation (set-theoretic or otherwise) has been left unspecified.
Definition. A category of categories is a category $\mathcal C$ such that
(i) each object of $\mathcal C$ is a category,
(ii) for any objects $A$ and $B$ of $\mathcal C$, the
morphisms from $A$ to $B$ in $\mathcal C$ are exactly the functors from $A$ to $B$,
(iii) for any object $A$ of $\mathcal C$, the identity morphism $1_A$ at $A$ is the identity functor on $A$,
(iv) the composition law for morphisms of $\mathcal C$ is given by
composition of functors.
Definition. A category $\mathcal C$ is universal if
(i) $\mathcal C$ is a category of categories,
(ii) every category is isomorphic to some category which
belongs to $\mathcal C$.
Theorem. There is no universal category.
Proof
In order to give the proof, we use the following definitions and lemmas.
Let $M$ be the category with exactly two objects $A$ and $B$ and exactly three morphisms $1_A$, $1_B$, and $f: A\to B$.
Let $N$ be the category with exactly three objects $A, B, C$ and exactly six morphisms $1_A, 1_B, 1_C, f: A \to B, g: B \to C, h: A \to C$.
Lemma. Let $\mathcal C_1$ and $\mathcal C_2$ be two isomorphic categories such that
(i) $\mathcal C_1$ is a category of categories,
(ii) $\mathcal C_2$ is a category of categories,
(iii) $\mathcal C_1$ contans categories that are isomorphic to $M$ and $N$.
Then every category belonging to $\mathcal C_1$ is isomorphic to some category belonging to $\mathcal C_2$, and vice versa.
Proof (sketch). If $C$ is any category belonging to $\mathcal C_1$, there is an
obvious one-to-one correspondence between morphisms of $C$ and functors
from $M$ into $C$, and compositions of such morphisms are picked out by
functors from $N$ into $C$.
Thus the isomorphism types of categories belonging to $\mathcal C_1$ are determined by the isomorphism type of $\mathcal C_1$ itself.
Furthermore, the same goes for $\mathcal C_2$, because $\mathcal C_2$ also contains categories
isomorphic to $M$ and $N$.
The lemma follows.
Definition. A category $\mathcal C$ is autistic if
(i) $\mathcal C$ is a catogory of categories,
(ii) $\mathcal C$ is isomorphic to some category which is an object
of $\mathcal C$.
$\ $ A category $\mathcal C$ is pseudoautistic if it is isomorphic to some autistic category.
Lemma. Let $\mathcal C$ be a category of categories such that
(i) $\mathcal C$ is pseudoautistic,
(ii) $\mathcal C$ contains categories isomorphic to $M$ and $N$.
Then $\mathcal C$ is autistic.
Proof. This follows immediately from the previous lemma.
To prove the theorem, let $\mathcal C$ be a universal category.
Let $\mathcal D$ be the full subcategory of $\mathcal C$ consisting of all categories belonging to $\mathcal C$
which are not pseudoautistic.
Then $\mathcal D$ is again a category of categories.
Also, since $M$ and $N$ are not pseudoautistic, $\mathcal D$ contains
categories isomorphic to $M$ and $N$.
Suppose first that $\mathcal D$ is autistic.
Then $\mathcal D$ is isomorphic to some category $E$ which belongs to $\mathcal D$.
Then $E$ is pseudoautistic. Hence, by definition of $\mathcal D$, $E$ does not belong to $\mathcal D$.
This is a contradiction.
We have shown that d is not autistic.
It follows by the previous lemma that $\mathcal D$ is not pseudoautistic.
Since $\mathcal C$ is universal, $\mathcal D$ is isomorphic to some category $E$ which belongs to $\mathcal C$.
Hence $E$ is not psuedoautistic. Hence $E$ belongs to $\mathcal D$.
Thus $\mathcal D$ is autistic. This is a contradiction.
The theorem is proved.
We write $\mathcal C(A,B)$ for the homset of arrows $A\to B$ in some category $\mathcal C$.
Let $\varphi:\mathcal C_1\to\mathcal C_2$ a category isomorphism.
Besides that functors from the two specific categories, 'the arrow' $M$ and 'the composition' $N$, encode the category structure, the main (implicit) statement is that $\varphi(M)$ must be isomorphic to $M$ and $\varphi(N)$ to $N$.
For any objects $C,X\in Ob\,\mathcal C_1$ we have $\ \mathcal C_1(X,C)\cong\mathcal C_2(\varphi(X),\,\varphi(C))$.
Moreover, if $C=X$, we get isomorphism of the endomorphism monoids.
Applying it to $X=C=M$ yields that $\varphi(M)$ has exactly 3 endofunctors, corresponding to the 3 endofunctors $M\to M$ (which themselves correspond to arrows of $M$), with the same monoid structure.
Note that a nontrivial category $C$ has at least one more endofunctors as objects because of the constant functors and the identity.
This itself shows that $\varphi(M)$ can have at most $2$ objects, and investigating endofunctors of categories with $1$ or $2$ objects leads to the more specific description of $\varphi(M)$.
Once $\varphi(M)\cong M$ is shown, based on this, one can also prove $\varphi(N)\cong N$, and then the rest follows.