In the category Sets consider the object (i.e., set) $A$ defined as the positive real numbers, another object (set) $B$ defined as the negative reals, and a function $f$ defined as the negative square root. So, $f(25)=-5$ and $f(9)=-3$ , etc.

Now define a functor $F$ from Sets to Sets such that $F(A)=-A$, $F(B)=-B$, and $F(f)=f$.

Strangely, we now we have a "function" that takes the negative square root of a negative number and gives a positive real number. This is clearly impossible in any normal sense of sets and functions.

As stated in Wikipedia, "the category Set is the category whose objects are Sets. The arrows or morphisms between sets $A$ and $B$ are all triples $(f,A,B) $ where $f$ is a function from $A$ to $B$."

Here is my problem: The functor above from Sets to Sets shows that f need NOT be a function from one set to another set (because the negative square root function cannot map a negative number to the positive reals). All this violates the definition of the category Sets, and/or the definition of a function, and/or the definition of a set (of numbers).

What am I doing wrong? Is $F$ not a functor? (And if not, how would I be able to tell it's not a valid functor?)

EDIT: Let me restate the second part of my question (in parentheses above), since it has not been answered so far: Suppose we have a given category $D$ and we are given a pair of functions (f,g) where the domain of f is the objects of $D$ and the domain of g is the arrows of $D$. This pair may, or may not, constitute a functor from $D$ to some other (unspecified) category.

My question concerns what methods/tests/procedures one can use to determine whether such a pair of functions constitutes a functor, or not. (Perhaps no general methods exist, or perhaps the question is too broad, but if that is the case, then that should be the answer to my question.)

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    $\begingroup$ What does $-A$ mean if $A$ is the set of Supreme Court justices? $\endgroup$ – John Douma Jul 11 '17 at 21:07
  • $\begingroup$ @JohnDouma $A$ is the set of real numbers so for $A$ his definition of $F$ is fine, it is tough, only a partial definition. You don't need the rest of the definition of $F$ to see what is wrong here tough (see my answer) $\endgroup$ – Jens Renders Jul 11 '17 at 21:11
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    $\begingroup$ To define a functor $F\colon Set\to Set$ you must specify for each set $S$ a corresponding set $F(S)$, and then similarly a correspondence on the morphisms. In your 'construction' of $F$ you did not specify a functor at all. Perhaps there lies your confusion. $\endgroup$ – Ittay Weiss Jul 11 '17 at 21:30
  • $\begingroup$ Please see detailed Edit and consider reopening. Thanks. $\endgroup$ – PossumP Jul 12 '17 at 21:40

$F$ is not a functor. By definition of a functor, $F(f)$ should be a map from $F(A)$ to $F(B)$ i.e. $$F(f): F(A) \rightarrow F(B)$$ $f:A\rightarrow B$ cleary doesn't satisfy this as $F(A) = -A \neq A$ and same for $B$, so you cannot have $F(f)=f$ if $F$ is a functor

  • $\begingroup$ I don't think that's a proof that F is not a functor. I think you've more or less just repeated my "objection" to F. $\endgroup$ – PossumP Jul 11 '17 at 21:25
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    $\begingroup$ @MPitts It certainly is a proof that $F$ is not a functor, because the demand that $F(f)$ should be a map from $F(A)$ to $F(B)$ is part of the definition of a functor. $F$ doesn't satisfy this, so by definition is not a functor. I've editted in a link to this definition $\endgroup$ – Jens Renders Jul 11 '17 at 21:34
  • $\begingroup$ I disagree. The definition you cite never says F(.) must be a map between F(A) and F(B). Furthermore, consider the dual of my example. In the dual we have f:B-->A which takes negative square roots of negative numbers (in B) to the positive real numbers (in A) - exactly the same operation as F(f):F(A)-->F(B). Yet that is valid in the dual category, so why is it not valid here? $\endgroup$ – PossumP Jul 11 '17 at 23:02
  • $\begingroup$ @MPitts Have you read it? I quote: "associates to each morphism $f:X \rightarrow Y$ in $C$ a morphism $F(f): F(X) \rightarrow F(Y) $ in $D$ such that..." $\endgroup$ – Jens Renders Jul 11 '17 at 23:07
  • $\begingroup$ @MPitts And your example in the dual category doesn't really make any sense. In the dual category of Set, the morfisms are no longer functions $\endgroup$ – Jens Renders Jul 11 '17 at 23:15

Here is a category D and a functor F that allows the kind of "nonsense" morphisms/functions I was trying to demonstrate in my example. (Thanks to Jens' comments I can see that Sets was not the right choice for the codomain since that implies morphisms must be functions.)

Let D be a 3 object category with A={25}, B={-5}, and C={-10}. We define f:A$\rightarrow$B where f is the negative square root function. Also g:B$\rightarrow$C where g(x)=2x.

Now consider the following functor that maps D to its dual: Let the functor F map A to C, B to B, C to A, f to g, and g to f. This means that in the dual, F(g)=f: -5 $\rightarrow$ 25. But by definition f is a function that takes the negative square of its argument, so how can f=F(g) take -5 to 25? Obviously this does not make sense in our "normal" way of thinking about functions.

This shows that F(g)=F(A)$\rightarrow$F(B) does not have to be "reasonable" because here we have f=F(g):-5 $\rightarrow$ 25.

  • $\begingroup$ $f$ is not the same thing as $F(f)$, so what's your point? $\endgroup$ – Alex Provost Jul 12 '17 at 1:56
  • $\begingroup$ @Alex, I've correct typo and clarified text. Please re-read and re-evaluate. $\endgroup$ – PossumP Jul 12 '17 at 4:34
  • $\begingroup$ The dual category has morphisms that go in the opposite direction. The morphisms $f:A \to B$ and $g:B \to C$ in $D$ are not morphisms in $D^{op}$, so $F(f)=g$,$F(g)=f$ don't make sense if $F$ is to be a functor $D \to D^{op}$. Anyway, the category doesn't "care" what your objects look like or how you call your morphisms, so I still don't understand the issue you're having. $\endgroup$ – Alex Provost Jul 12 '17 at 17:16
  • $\begingroup$ @Alex, I don't think you're reading my post carefully. Agreed, f:A-->B does not exist or make sense in the dual. But the functor maps f to g and A to C and B to B, so we get g:C-->B as desired for the dual category. Similarly the functor maps g:B-->C to f:B-->A as desired. (See Math.stackexchange article 2350047 which addresses this exact functor.) If you don't accept my isomorphic functor between D and its dual, maybe it would be useful to indicate what do you think is the proper isomorphic functor that transforms D to its dual? $\endgroup$ – PossumP Jul 12 '17 at 21:30
  • $\begingroup$ $g$ is already defined: it is a morphism $B \to C$ in $D$. This $g$ is not the same thing as the corresponding morphism $C \to B$ in the opposite category. (You might want to call this morphism $g^{op}$.) Maybe you want to say that $F(f) = g^{op}$? Anyway, the duality functor one usually uses maps any object to itself and sends any morphism $f$ to its opposite $f^{op}$. However this is generally not an isomorphism: for instance, the category of sets isn't isomorphic to its dual (think about initial and terminal objects there). $\endgroup$ – Alex Provost Jul 13 '17 at 3:17

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