Julio wrote:

> I have been confused by the 'op' thing for quite a while.

Good! This will build up the desire in you to do lots of calculations, to resolve this confusion.

For starters, do you understand what the category \\(\mathcal{C}^{\mathbf{op}}\\) is?

If so, the next step is to understand why there is no functor

\[ \mathrm{hom} : \mathcal{C} \times \mathcal{C} \to \mathrm{Set} \]

sending each pair of objects \\( (c,c') \\) to the 'homset' \\(\mathcal{C}(c,c')\\) - that is, the set of morphisms from \\(c\\) to \\(c'\\).

The best possible way is for you to try to define this (nonexistent) functor and see what goes wrong. I've told you what it does to an object of \\( \mathcal{C} \times \mathcal{C} \\) - that is, a pair \\( (c,c')\\) of objects in \\(\mathcal{C}\\). So your job is to define what this functor does to _morphisms_.

In other words, given

\[ (f,f'): (c,c') \to (d,d') ,\]

try to cook up a map from \\(\mathcal{C}(c,c')\\) to \\(\mathcal{C}(d,d')\\).

When you try this, and see exactly why it doesn't work, you'll be ready to see why we _do_ have a functor

\[ \mathrm{hom} : \mathcal{C}^{\mathrm{op}} \times \mathcal{C} \to \mathrm{Set} \]

You can then define what this does to morphisms, and show that it _does_ work.

There's really nothing better, when learning math, than just doing calculatiions and seeing what happens. _Watching_ people do math is like watching the Olympics. It's fun, but it doesn't make you able to run faster. In the sort of category theory we're doing now, the calculations are all really easy. It just takes a bit of courage to dive and try them.

But fear not! I _will_ explain the hom-functor

\[ \mathrm{hom} : \mathcal{C}^{\mathrm{op}} \times \mathcal{C} \to \mathrm{Set} \]

I _need_ to explain it, to really explain adjoint functors!

But I decided to try, at first, to give people a bit of intuition about adjoint functors

before delving into this matter.

> I have been confused by the 'op' thing for quite a while.

Good! This will build up the desire in you to do lots of calculations, to resolve this confusion.

For starters, do you understand what the category \\(\mathcal{C}^{\mathbf{op}}\\) is?

If so, the next step is to understand why there is no functor

\[ \mathrm{hom} : \mathcal{C} \times \mathcal{C} \to \mathrm{Set} \]

sending each pair of objects \\( (c,c') \\) to the 'homset' \\(\mathcal{C}(c,c')\\) - that is, the set of morphisms from \\(c\\) to \\(c'\\).

The best possible way is for you to try to define this (nonexistent) functor and see what goes wrong. I've told you what it does to an object of \\( \mathcal{C} \times \mathcal{C} \\) - that is, a pair \\( (c,c')\\) of objects in \\(\mathcal{C}\\). So your job is to define what this functor does to _morphisms_.

In other words, given

\[ (f,f'): (c,c') \to (d,d') ,\]

try to cook up a map from \\(\mathcal{C}(c,c')\\) to \\(\mathcal{C}(d,d')\\).

When you try this, and see exactly why it doesn't work, you'll be ready to see why we _do_ have a functor

\[ \mathrm{hom} : \mathcal{C}^{\mathrm{op}} \times \mathcal{C} \to \mathrm{Set} \]

You can then define what this does to morphisms, and show that it _does_ work.

There's really nothing better, when learning math, than just doing calculatiions and seeing what happens. _Watching_ people do math is like watching the Olympics. It's fun, but it doesn't make you able to run faster. In the sort of category theory we're doing now, the calculations are all really easy. It just takes a bit of courage to dive and try them.

But fear not! I _will_ explain the hom-functor

\[ \mathrm{hom} : \mathcal{C}^{\mathrm{op}} \times \mathcal{C} \to \mathrm{Set} \]

I _need_ to explain it, to really explain adjoint functors!

But I decided to try, at first, to give people a bit of intuition about adjoint functors

before delving into this matter.