So, despite the opposite approach on the second page.

In the first task monomorphisms are characterized again, as we have already seen for other categories.

And to characterize isomorphisms means to give a reasonable description, which of course should not be the same as the definition of a monomorphism.

Formally this task is completely under-specified.

Of course you can take the original definition or shape it trivially or something like that.

That is not meant, but meant is something like what we have seen in the category of quantities.

For example an injective image or whatever comes out of it.

I will not reveal what comes out of it.

And just as the isomorphisms.

We will understand task B at the end of today's session, where we introduce the products.

Eventually task two will be understood, if you do not understand it immediately after today's session.

After the next time, where we will show commutativity and neutral element properties for the product on Friday.

That we have the associativity left, which is then for a note.

Yes, task three.

We introduce what products are today.

And then this little statement about the preservation of monomorphisms under products is shown.

It is clear on quantities.

We can read that on quantities. We have already introduced the product on quantities.

So it is not surprising.

Monomorphism means injective.

I have to show that if f and g are injective, then f cross g is also injective.

That is a one-digit that I prove once.

If I have a pair, then I have two pairs.

If I start on the first component g and the second component f and the second component g,

then the elements are thrown in the same way.

Then the components are thrown in the same way.

And by injectivity of these two representations, then both components are equal.

So the pairs are equal.

This very simple argument is categorized here.

We see that this argument survives under the categorical definition.

And then task four again.

Dual statement comes today, which means duality.

So what comes out when I turn all the arrows?

And then here task four.

Here a stricter term of monomorphism is introduced.

Here it is strictly not shown that it is a monomorphism.

So you could also do the task of making each such cut a monomorphism.

Well, we don't want to do that here.

But we want to prove an statement about such cuts.

Cuts are practically morphisms that have a left inverse.

If we remember the definition of isomorphism, then the inverse is that we require both sides.

So there must be identity when I compose the left edge and when I compose the right edge.

And if I only demand that on the left side, then such a thing is called a cut.

And if I demand it on the right side, it is called a retraction.

That somehow went to float when editing too quickly.

That was there.

That's probably commented out now or something.

Good, I'll add that again.

So what comes in the second task with the retractions is the dual to the cut.

And then there is again such a small lemma to prove.