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.
Thanks in advance.
Best Answer
Consider the following property of a category $\mathcal C$:
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).