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This is a followup to this question. (Matt Feller also mentioned this followup in a comment to the question linked to above.)

For any category $\mathcal C$ write $\mathcal C^{\mathcal C}$ for the category of endofunctors of $\mathcal C$, and write $[\operatorname{Ob}({\mathcal C})]$ for the collection of isomorphism classes of objects of $\mathcal C$. (Note that $[\operatorname{Ob}({\mathcal C})]$ is not necessarily a set.)

Let $\mathcal C$ be a category which is not equivalent to a category having exactly one object and one morphism.

Are the following statements necessarily true?

(1) There is no surjection $[\operatorname{Ob}(\mathcal C)]\to[\operatorname{Ob}(\mathcal C^{\mathcal C})]$.

(2) There is no injection $[\operatorname{Ob}(\mathcal C^{\mathcal C})]\to[\operatorname{Ob}(\mathcal C)]$.

A positive answer to at least one of the above questions would also answer the question linked to above. A negative answer to at least one of the above questions would also answer this older question. [Emil Jeřábek noticed this mistake.]

(We assume that we are working in ZFC.)

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  • $\begingroup$ If we're working in ZFC, then (1) and (2) are equivalent. For that matter, technically we need to be working in some sort of theory that allows us to talk about classes, anyway. $\endgroup$
    – Tim Campion
    Commented Jun 25, 2019 at 13:28
  • $\begingroup$ @Tim: Why do you think they are equivalent? (I'm not saying they are not, I just want to see if your logic is correct. To wit, it is consistent with ZFC that there is a class that surjects onto the ordinals, but there is no injection from the ordinals into it.) $\endgroup$
    – Asaf Karagila ♦
    Commented Jun 25, 2019 at 13:35
  • $\begingroup$ @AsafKaragila Yeah, I guess I was thinking "If we formalize this in ZFC + a universe rather than a class theory, then they're equivalent." But I didn't get the impression that the question is supposed to be about the subtelties of classes in ZFC. $\endgroup$
    – Tim Campion
    Commented Jun 25, 2019 at 13:38
  • $\begingroup$ @Tim: Of course, but it might just end up as one. $\endgroup$
    – Asaf Karagila ♦
    Commented Jun 25, 2019 at 13:39
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    $\begingroup$ You just gave a negative answer, but as far as I can tell, this does not resolve the linked question on poset isomorphic to their endomorphism posets. (The endomorphism poset of $\mathbb R$ is very different from $\mathbb R$.) So I believe that claim in the question is faulty. $\endgroup$
    – Emil Jeřábek
    Commented Jun 27, 2019 at 15:13

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If $\mathcal C$ is the category attached to the ordered set $(\mathbb R,\le)$, then $[\operatorname{Ob}(\mathcal C)]$ coincides with the set $\mathbb R$ and $[\operatorname{Ob}(\mathcal C^{\mathcal C})]$ coincides with the set of weakly increasing functions from $\mathbb R$ to $\mathbb R$. But it is well known that these two sets are equipotent: see for instance this answer of Brian M. Scott or this answer of mechanodroid.

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    $\begingroup$ Interesting! And it seems that in this example, $|\text{Mor}(\mathcal{C})|=|\text{Mor}(\mathcal{C}^{\mathcal{C}})|$ as well. (In each case, we have the identity morphism for each object, so at least $2^{\aleph_0}$ morphisms, but since they are both posets, we also have no more than $|\text{Ob}\times\text{Ob}|=2^{\aleph_0}\cdot 2^{\aleph_0}=2^{\aleph_0}$ morphisms.) $\endgroup$
    – Matt Feller
    Commented Jul 17, 2019 at 22:28

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