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This is going to be a bit of a long post, so fair warning.

So from what I could gather from some research, on the first link below, it goes that the category $Set$ is not isomorphic to its opposite, $Set^{op}$. My issues concern proving why.

One of the answers seems to make use of properties regarding the empty set. Namely, that, for all objects $X \in Set$, the arrow $X \rightarrow \emptyset$ exists if and only if $X = \emptyset$. This arrow would be isomorphic, clearly. There also exists an arrow $\emptyset \rightarrow X$ for all objects $X \in Set$.

So in $Set^{op}$, the dual idea reverses these arrows, i.e. for all $X$ we have an arrow $X \rightarrow \emptyset$. Apparently from what I could find at the second link below, this isn't an inherent violation ... for some technical reasons that I don't really understand. But I guess I'm basically mean to take the "reversing all arrows" part of the definition of opposite category very literally, and that it doesn't inherently break any rules by itself.

So my question is, then, how exactly do these become relevant to proving $Set$ is not isomorphic to its opposite? Because if we don't really have these violations, it seems trivial to define a functor $T$ such that

  • Object function: $X \mapsto T(X)$ as according to the below
  • Arrow function: $(f : A \rightarrow B) \mapsto (T(f) : T(A) \rightarrow T(B))$, where $T(A) = B$ and $T(B) = A$.

But obviously that's not right since it doesn't really invoke any assumptions about $Set$ and could be generalized to prove $\mathscr{C} \cong \mathscr{C}^{op}$ for any category $\mathscr{C}$.

Supposedly, according to a third post, this has something to do with the notion of "categorical properties." Namely, that - whatever these properties are, that term hasn't come up in my course at all - these are preserved by functors - which wasn't even mentioned when we covered functors.

So I'm trying to rationalize my way through this, but I can't be sure because ... well, I can't really figure out what a categorical property is, and how that applies in this case.

Does anyone have any nudges in the right direction for me?


Link 1 - Why is every category not isomorphic to its opposite?

Link 2 - What is the opposite category of $Set$?

Link 3 - Proof that $ \textbf{Set}\ $ is not isomorphic to its dual

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    $\begingroup$ Your "definition" of $T$ isn't a defintion at all. You haven't actually said what $T(f)$ is, and moreover it does not make sense to say $T(A)=B$ and $T(B)=A$ because what if there are maps $f:A\to B$ and $g:A\to C$ for two different objects $B$ and $C$? Then you would have to have both $T(A)=B$ and $T(A)=C$. $\endgroup$ – Eric Wofsey Oct 24 '18 at 3:36
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    $\begingroup$ In any case, I would suggest that you not think in terms of functors but instead just think in terms of the intuitive notion of isomorphism, which is exactly the same for categories as it is for any other kind of algebraic structure. Two categories are isomorphic if they are "the same" except that the elements (objects and morphisms) have been given different names. "Categorical properties" are just properties that stay the same when you rename the elements. This is just like how, for instance, being abelian is a property of groups that is preserved by isomorphisms of groups. $\endgroup$ – Eric Wofsey Oct 24 '18 at 3:41
  • $\begingroup$ The root idea I was trying to basically get at is that it would map each arrow in $Set$ to its reversal in $Set^{op}$. I'm well aware that it's wrong as an approach in the first place, even if it's for other reasons, as I acknowledge later. $\endgroup$ – Eevee Trainer Oct 24 '18 at 3:41
  • $\begingroup$ I also don't understand that definition of categorical property either. "Renaming" invokes the idea of just naming variables different things, and I imagine that's not at all what you're trying to get at. $\endgroup$ – Eevee Trainer Oct 24 '18 at 3:51
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    $\begingroup$ Well, what intuition do you have for what it means for two groups to be isomorphic? For instance, can you understand immediately that if $G$ is a group which has an element of order $2$ and $H$ is a group which has no element of order $2$, then $G$ and $H$ cannot be isomorphic? That would be an example of a "group-theoretic property". $\endgroup$ – Eric Wofsey Oct 24 '18 at 3:52
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Let me flesh out Andreas Blass's answer at one of the questions you linked to which you seem to be misunderstanding.

Perhaps the easiest way to see that the category of sets isn't isomorphic to its dual is to observe that there is an object, namely the empty set, such that all morphisms into it are isomorphisms, but there is no object such that all morphisms out of it are isomorphisms.

What this is saying is that in $Set$ there exists an object $A$ (namely $\emptyset$) with the property that for every object $B$ and every morphism $f:B\to A$, $f$ is an isomorphism. The claim is that this is a categorical property (preserved by isomorphisms of categories*): if a category $\mathcal{C}$ has such an object $A$ and $F:\mathcal{C}\to\mathcal{D}$ is an isomorphism of categories, then $\mathcal{D}$ also has such an object.

The proof is pretty trivial: just observe that $F(A)$ will have the same property in $\mathcal{D}$. Indeed, suppose $B$ is an object of $\mathcal{D}$ and $f:B\to F(A)$ is a morphism. Then $F^{-1}(f):F^{-1}(B)\to A$ is a morphism, so by the assumed property of $A$ it is an isomorphism. Thus $f=F(F^{-1}(f))$ is also an isomorphism, as desired.

What's going on here is that we can freely "translate" between $\mathcal{C}$ and $\mathcal{D}$ by applying $F$ and its inverse, so any reasonable property of an object $A$ in $\mathcal{C}$ will also be true of the object $F(A)$ in $\mathcal{D}$.

OK, now let's get back to our story of $Set$. In the category $Set$, there is an object $A$ with the property stated above (namely $A=\emptyset$). If $Set$ and $Set^{op}$ were isomorphic, then there would exist some object $A'$ with the same property in $Set^{op}$. Let's unravel what this means. We have an object $A'$ of $Set^{op}$ (i.e., a set) such that for any object $B$ of $Set^{op}$ (i.e., a set) and any morphism $f:B\to A'$ (i.e., a function $f:A'\to B$), $f$ is an isomorphism (i.e., $f$ is a bijection). Translating this back into ordinary statements about sets as indicated in parentheses, we have a set $A'$ such that for any set $B$, every function $f:A'\to B$ is a bijection. However, no such set exists: for instance, if $A'$ is any set and $x$ is not an element of $A'$, then $B=A'\cup\{x\}$ is a set and the inclusion map $f:A'\to B$ is not a bijection.

So, no such object $A'$ of $Set^{op}$ exists, and so $Set^{op}$ is not isomorphic to $Set$.

*Actually, usually when people talk about "categorical properties", they really mean properties that are preserved by equivalences of categories, not just isomorphisms. That is a more subtle story that I won't get into here, since your question is just about isomorphisms.

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  • $\begingroup$ Okay, I just want to check my understanding of your argument; not sure if people who answer questions get notifications of other answers (still kinda new here), so just for the sake of clarifying I added it as a separate answer due to length. $\endgroup$ – Eevee Trainer Oct 24 '18 at 4:34
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Okay, I just want to ensure that I understand the argument. (Just adding this as an answer because of length.)

First, on categorical properties: if some collection of objects and morphisms in some category $\mathscr{C}$ exhibit some sort of "unusual" property - as the empty set does here with every map into it being an iso - then any category $\mathscr{D}$ where $\mathscr{C} \cong \mathscr{D}$ must also have some collection of objects and morphisms exhibiting this same property.

(A sticking point is "unusual:" analogizing categorical properties to structure in an intuitive sense is basically what I'm doing. That would be to say, isomorphisms of categories would preserve this "structure," or as I'm referring to it, "unusual property." I'm not 100% sure when this can necessarily be invoked or used - if only certain things can be preserved like this for example.)

In this case, since there exists an object in $Set$ such that every arrow to it is an isomorphism, then, if $Set \cong Set^{op}$, there must exist another such object in $Set^{op}$. (The fact that it is only one such object and map - the empty set and its arrow onto itself - is just a technicality and doesn't really matter in the grand scheme of things. This was another sticking point.)

So we suppose such an object $A$ exists in $Set^{op}$. Then, for all objects $B$ and arrows $f : B \rightarrow A$, $f$ is an isomorphism. This would correspond to a function $f : A \rightarrow B$ since we're in the opposite category and arrows are reversed.However, we can construct $B$ such that $f$ (in this latter context) being iso is impossible: if $x \notin A$ and $B = A \cup \{x\}$, then the map is clearly not a bijection.

Thus, since $f : A \rightarrow B$ can't be iso for all $B$, $f : B \rightarrow A$ in $Set^{op}$ can't be an iso for all $B$. That is, there does not exist an object such that all maps to it are iso in $Set^{op}$, where $\emptyset$ is such an object in $Set$. Thus, since this property isn't exhibited in one when it is in the other, the two categories cannot be isomorphic.

Is that at least a rough approximation of what you were getting at earlier? @EricWofsey

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    $\begingroup$ Yeah, this is the right idea. Categorical properties are more general, though: they could be properties of all objects rather than just some special objects, for instance. Categorical properties are just any properties that can be "transferred" from one category to another using an isomorphism and its inverse, like the argument I gave in my answer. With a bit of experience, you will eventually be able to instantly recognize such properties instead of having to write out the argument in each case. $\endgroup$ – Eric Wofsey Oct 24 '18 at 4:52

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