Constructing Dual notions via Duality

Question

First of all, I do not have much mathematical background and I have minimal category theory knowledge. I am just trying to understand one or two things about category theory because the concept sounds interesting. Please do not give answers like those in MacLane's book i.e answers for the working mathematician.

I understand that in Dual categories, all morphisms are essentially reversed.

For example having this Pullback in the category of sets, will reversing $p_1$ and $p_2$ leaving all the elements inside the set as they are and the functions represented by the morphisms $p_1$,$p_2$ also as is but just switching the domain and codomain make it a pushout?

Answer 1

A pullback is not just the diagram. It must also satisfy some properties:

1. The diagram must commute, in other words $f\circ p_1=g\circ p_2$.

2. [Universal property] $P$ must be such an object, that for any object C of the category, if there are morphisms $C\to X$ and $C\to Y$, then it must also have a unique morphism $C\to P$ and these morphisms must commute, again.

So, for the definition of a pushout, you have to reverse the direction of the morphisms not only at the diagram, but also at the properties. So, they become like this:

1. The (reversed) diagram must commute, in other words $p_1\circ f=p_2\circ g$.

2. [Universal property] $P$ must be such an object, that for any object C of the category, if there are morphisms $X\to C$ and $Y\to C$, then it must also have a unique morphism $P\to C$ and these morphisms must commute, again.

Answer 2

Adding to @frabala 's answer:

A pullback in category $C$ corresponds to a pushout in the category $C^{OP}$, so your pullback in $Set$ corresponds to a pushout in $Set^{OP}$. However the morphisms in $Set^{OP}$ are not functions. They are just formal arrows, so working in $Set^{OP}$ is not very intuitive.

Now, I warn you that some working mathematician might casually mention that $Set^{OP}$ is equivalent to the category of complete atomic boolean algebras ($CABA$), so some meaningful interpretation in $Set^{OP}$ is still possible, but this might sound like an answer from Mac Lane's book, and I won't do that :-)

By the way, welcome to Math.SE!