# Why is $Set$ not equivalent to $Set^{op}$?

I understand this question already has an answer on here, but I do not understand how the answer justifies that $Set$ not equivalent to $Set^{op}$.

The explanation is that if they were equivalent and if the terminal object in $Set$ has an equivalence invariant property, then the initial object has the dual property.

The terminal objects are obviously singleton sets, and the initial object is the empty set. I can see that every map into the empty set is an isomorphism, and that there exists maps $\{a\}\to X$ which are not invertible, but how does this relate to the non-equivalence of $Set$ and $Set^{op}$?

If someone could elaborate further I would greatly appreciate it.

• The second paragraph is the explanation. Try dualising the two properties in question. Mar 7, 2016 at 10:42
• It is even easier to observe that $\mathbf{Set}$ has an arrow from a terminal object to a non-terminal one, but $\mathbf{Set}^{\rm op}$ does not (such an arrow would a function from a nonempty set to the empty set). Mar 7, 2016 at 10:46
• But why is the explanation in the second paragraph true? It just seems odd to show a nonequivalence f a category and its opposite by just talking about one. Mar 7, 2016 at 10:46
• The part as to why we can make this conclusion without even considering $Set^{op}$. Mar 7, 2016 at 10:49
• @BrandonThomasVanOver: No, in the comment above I'm only mentioning terminal objects. The terminal objects in $\mathbf{Set}$ are the singletons, but the terminal object in $\mathbf{Set}^{\rm op}$ is the empty set. So I'm not speaking about singletons in the context of $\mathbf{Set}^{\rm op}$ at all. Mar 7, 2016 at 10:54

Consider the following property of a category $\mathcal C$:

If $f: A\to B$ is a morphism and $B$ is an initial object, then $f$ is an isomorphism.

Since this property is stated only in terms of existence properties of certain morphisms (and does not depend on uniqueness of objects even if we unfold the definition of an initial object), if we have two equivalent categories they will either both have this property or both not have this property.

The explanation you reproduce consists of the observation that $\mathbf{Set}$ has this property (because the only morphism into an initial object is the empty function $\varnothing\to\varnothing$, which is an isomorphism), whereas $\mathbf{Set}^{\rm op}$ does not (there are functions with singleton domains that are not bijections, and such a function is in $\mathbf{Set}^{\rm op}$ a non-iso morphism into an initial object).

• So from my understand the singleton sets become initial in the opposite category, so why isn't it enough to say that not every map into a singleton is an isomorphism then? Mar 7, 2016 at 11:07
• @BrandonThomasVanOver: You're confusing the directions of morphisms and map. Remember that in $\mathbb{Set}^{\rm op}$ a morphism from $A$ to $B$ is a map from $B$ to $A$. So "morphism into an initial object" in $\mathbf{Set}^{\rm op}$ is the same as "map from a singleton". Mar 7, 2016 at 11:31
• @HenningMakholm, the homs in $Set^{op}$ are not maps (functions). You mention in your comment to the OP above that "such an arrow would [be] a function...". Jul 29, 2019 at 7:24

Statement 1 : Let $C\simeq D$ be two categories which are equivalent thanks to a functor $F : C \to D$. If $X$ is an initial object in $C$ then $F(X)$ is an initial object in $D$.

Statement 2 : If $X$ is an initial object in $C$ (for some category $C$), it is a terminal object in $C^{op}.$

Statement 3 : If $F : C \to D$ is an equivalence between categories with products, one has $F(X \times Y) \simeq F(X) \times F(Y)$ for all $X,Y$.

Now suppose to have a functor $F : Set \to Set^{op}$ which gives an equivalence between them. Then, because $\emptyset$ is initial in $Set^{op}$, we get that $F(\emptyset)$ is initial in $Set^{op}$. (Statement 1). After that, we get that $F(\emptyset)$ is terminal in $Set$. (Statement 2). Hence, there is $a$ such that. $F(\emptyset) = \{a\} .$ Now apply statement 3 $$\{a\} = F(\emptyset) = F(\emptyset \times \emptyset) \simeq F(\emptyset) \times F(\emptyset) = \{a\} \bigsqcup \{a\}.$$ So you have established a bijection between $\{a\}$ and $\{a\} \bigsqcup \{a\},$ which leads to a contradiction.

• The function $f: \{a\}\to\{a\}\times\{a\}$ given by $f(a) = \langle a,a\rangle$ is a bijection. (However, the categorical product in $\mathbf{Set}^{\rm op}$ is of course disjoint union, so what you actually need is a bijection between $\{a\}$ and $\{1\}\times\{a\}\cup\{2\}\times\{a\}$ which is still impossible). Mar 7, 2016 at 12:32
• Of course, corrected. Mar 7, 2016 at 13:25

The opposite to Set is the category of Complete Atomic Boolean Algebras, see

https://ncatlab.org/nlab/show/Set#OppositeCategory