Category Theory - 2018-07-01

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user131753

4:00 PM

Hello @LeakyNun.

user131753

I am having a problem in Category Theory regarding the definition of Dual Category. Would you mind to discuss that with me@LeakyNun?

Leaky Nun

depends on what the problem is

user131753

@LeakyNun Ok. Fair enough.

user131753

The basic problem in the definition of Dual Category of a given category $\mathbf{A}$ is that: Why do we denote the morphisms in the dual of $\mathbf{A}$ (henceforth $\mathbf{A}^{\text{op}}$ by the same "letters")?

Leaky Nun

because they're the same

it's just the composition that has changed

@user170039 recall that a category consists of a set of objects Ob(A), a set of morphisms Mor(X,Y) between each two objects X and Y, and a composition function Mor(Y,Z) x Mor(X,Y) -> Mor(X,Z)

user131753

4:07 PM

@LeakyNun In what sense exactly?

user131753

@LeakyNun Not really. Direction of the morphisms has also changed.

Leaky Nun

Let's denote it as $\langle Ob(A), Mor_A(X,Y), Comp_A(X,Y,Z) \rangle$

Then, $Mor_{A^{op}}(X,Y) := Mor_A(Y,X)$

you don't really need a new letter

for the elements inside

if $f \in Mor_A(Y,X)$, then $f \in Mor_{A^{op}}(X,Y)$

it's the same $f$

this is working within ZFC

user131753

@LeakyNun In what sense of "sameness" exactly?

Leaky Nun

they are equal as objects in ZFC

user131753

@LeakyNun You mean equal by Axiom of Extension, right?

Leaky Nun

4:12 PM

extension is a way of proving equality

"extensionally equal" doesn't make any sense

but in the informal meaning

no, they are definitionally equal

$Mor_{A^{op}}(X,Y)$ by definition contains every object in $Mor_A(Y,X)$

so if $f \in Mor_A(Y,X)$, then $f \in Mor_{A^{op}}(X,Y)$

@user170039 doesn't make any sense, unless you're throwing out the axiom of extension

user131753

@LeakyNun Can you clarify?

Leaky Nun

just ignore that

1 min ago, by Leaky Nun

$Mor_{A^{op}}(X,Y)$ by definition contains every object in $Mor_A(Y,X)$

52 secs ago, by Leaky Nun

so if $f \in Mor_A(Y,X)$, then $f \in Mor_{A^{op}}(X,Y)$

user131753

@LeakyNun What is "by definition" here?

Leaky Nun

by definition of $Mor_{A^{op}}(X,Y)$

7 mins ago, by Leaky Nun

Then, $Mor_{A^{op}}(X,Y) := Mor_A(Y,X)$

user131753

@LeakyNun I see.

user131753

4:20 PM

In fact, I was kind of wondering whether a different definition is possible and the following is a sketch of ideas that I could come up with. I will be glad to have your feedback regarding it @LeakyNun.

user131753

>**Definition 1.** Let $\mathbf{A}$ and $\mathbf{B}$ be two categories. A functor $\mathscr{D}:\mathbf{A}\to\mathbf{B}$ is said to be a *dual functor from $\mathbf{A}$ to $\mathbf{B}$* iff,

>- $\mathscr{D}_{\text{ob}}:\operatorname{Object}(\mathbf{A})\to\operatorname{Object}(\mathbf{B})$ is a bijection.

>- $\mathscr{D}_{\text{mor}}:\operatorname{Morphism}(\mathbf{A})\to\operatorname{Morphism}(\mathbf{B})$ is a bijection.

>- For all objects $A,B\in \operatorname{Object}(\mathbf{A})$, $${\mathscr{D}}{\large{|}_{\textsf{Hom}_{\mathbf{A}}(A,B)}}:\textsf{Hom}_{\mathbf{A}}(A,B)\to\textsf{Hom}_…

Leaky Nun

makes no sense at all to me

user131753

@LeakyNun Where exactly?

Leaky Nun

line 3

user131753

@LeakyNun "$\mathscr{D}_{\text{mor}}:\operatorname{Morphism}(\mathbf{A})\to\operatorname{M‌orphism}(\mathbf{B})$ is a bijection." - this one?

Leaky Nun

4:25 PM

right

user131753