7
$\begingroup$

An exponential object $B^{A}$ is defined to be the representing object of the functor $$\mathcal{C}\left(- \times A,B\right): \mathcal{C} \rightarrow Set$$ or equivalently, as the terminal object of $\left(-\times A \downarrow B\right)$. The dual concept is of the co-exponential object which is the initial object of the $\left(B\downarrow -\times A \right)$.

Is co-exponential object as useful as exponential object? What is the notation for them and what are the interesting examples of co-exponential objects? What is right (or left) adjoint of the functor which send any object to the co-exponential (with a fixed base)?

Thanks

$\endgroup$
3
  • 3
    $\begingroup$ Please, look at the notion of an "co-exponential object" in some special cases, and then generalize to arbitrary categories. You will see that this notion is not useful at all. You should do this everytime you play around with definitions: Examples! $\endgroup$ – Martin Brandenburg Dec 27 '13 at 2:15
  • 1
    $\begingroup$ @MartinBrandenburg On the contrary, I found many categories with interesting coexponentials: $Set^{op}, Top^{op}$, etc... . I also notice composition actually works left to right in these categories. $\endgroup$ – PyRulez May 10 '15 at 20:35
  • 4
    $\begingroup$ Really? Have you dualized products to coproducts? $\endgroup$ – Martin Brandenburg May 11 '15 at 10:18
4
$\begingroup$

If there is "co-exponential" $f:B\to C\times A$ as you suggest, let $f_1:B\to C$ and $f_2:B\to A$ be the compositions of $f$ with the projections. The universal property of $f$ says that given any $g=(g_1,g_2):B\to D\times A$, there is a unique $h:C\to D$ such that $g=(h\times 1)f$. Composing with the projections, $g=(h\times 1)f$ just says that $g_1=hf_1$ and $g_2=f_2$. So in particular, this means that given any map $g_2:B\to A$, $g_2=f_2$, i.e. $f_2$ is the unique map $B\to A$. It also means that every map from $B$ to another object $D$ factors uniquely through $f_1$, which by Yoneda says that $f_1$ is an isomorphism.

So to sum up, there can only exist a "coexponential" for $B$ and $A$ if there is exactly one map $B\to A$, in which case up to isomorphism the coexponential must be the map $B\to B\times A$ whose first coordinate is the identity and whose second coordinate is the unique map. Conversely, if there is exactly one map $B\to A$, the argument of the previous paragraph is reversible and shows that this map $B\to B\times A$ is indeed a "coexponential".


Beware, though, that your definition is not what people usually mean by "coexponential". The more usual definition would turn around all the arrows, including turning the product into a coproduct, so it would be an initial object of $(B\downarrow - \amalg A)$. (By this definition, a coexponential object in $\mathcal{C}$ would be the same thing as an exponential object in $\mathcal{C}^{op}$.)

$\endgroup$
3
  • 1
    $\begingroup$ Wouldn't there be a more sensible notion involving an arrow into a coproduct, rather than a product? $\endgroup$ – MJD Apr 15 '20 at 12:38
  • 1
    $\begingroup$ Yes, that is what "coexponential" usually means. $\endgroup$ – Eric Wofsey Apr 15 '20 at 14:44
  • 1
    $\begingroup$ I don't see any hint of that in your answer though. Maybe OP would appreciate having it pointed out that they might not have been asking the question they meant to? $\endgroup$ – MJD Apr 15 '20 at 20:15
2
$\begingroup$

I have given an explanation with example and a lot of references about how to think of co-exponentials. I have not yet worked out how useful they are for myself under that name.

But a very powerful argument to take them seriously is that reasoning requires the ability to think of how to falsify arguments put forward by someone else. So when someone puts forward an argument in constructive logic to the effect that $\Delta \vdash \alpha$ then it can be questioned by refuting $\alpha$, which in case one accepts the sequent requires one to find one or more statements in $\Gamma$ that one rejects also. This falsificationist reasoning is co-constructive, and that is where co-implications appear too, which is another word for co-exponential.

It has been forcefully argued that Scientific Reasoning is falsificationist by Karl Popper. In which case it would be very important.

$\endgroup$
3
  • 2
    $\begingroup$ You are talking about co-exponentials as the left adjoint of the coproduct, which was not the subject of this question. $\endgroup$ – Kevin Arlin Apr 15 '20 at 5:37
  • $\begingroup$ Ah. Thanks for making me aware of the distinction. I was thinking co-exponential referred only to co-implication. I'll need to think about this other concept then. $\endgroup$ – Henry Story Apr 15 '20 at 5:49
  • 2
    $\begingroup$ I do think it usually refers to the concept you had in mind. As you can see from Eric’s answer, the concept at issue in this question isn’t of any real interest. $\endgroup$ – Kevin Arlin Apr 15 '20 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.