I have the task one.

It's a bit longer.

Not anymore.

Task one.

It was about looking at the regular monomorphisms in two categories.

I'll draw the diagram for you.

I took two graphs.

In between we have a pair of morphisms.

As you all know, an equalizer is this pair with the object and the diagram.

So that if we have another object and from here this morphism...

The equalizer has to fulfill...

f eq equals f... no, g eq.

And if c has this property with fc equals gc,

then there has to be a single morphism h that equals this.

So, now I've thought about what we have and what I can do.

First of all, the morphisms are actually diagrams between quantities.

That means I get the idea that the eq equation has to be injective.

But if I assume that it is not an injective equation,

then I can't produce a single morphism here,

because I can choose any model for the corresponding ones that are depicted.

So, I'll write something like...

Implements of eq equals injective.

And that's why I thought I'd write the equalizer as...

or equalizer, now I called it eq...

as a connection of an isomorphism and then...

but not first...

an isomorphism and then an embedding or an inclusion.

And since I have to show something other than an isomorphism in the task position,

I can also give this one to myself and then consider h as a fraction of the knot quantity.

And that's why I changed my diagram to h' and my inclusion,

where I actually write the inclusion...

I'll write inc so that you can distinguish it a bit from c, because that's not possible.

Now we've thought about it, we have the injective representation.

And now I want to show that...

once h is a subgroup, I can then specify h.

By choosing a few fg for which h is the equalizer.

So I say h... I'll write it like this...

g1 subgroup...

and then I need f,g from hom, grad, g1, g2,

so that the quantity fx equals gx is exactly h.

And I mean... well...

if g2 doesn't allow it...

I don't know, can I also choose g2?

Yes.

So then g2 could also be a subgroup and a g1, right?

Can I say that?

No.

You don't have to?

Maybe.

You have to get the f and g so that they keep the edges.