MHB Category Theory - Initial and Final Objects

[Math Amateur]

I am reading Paolo Aluffi's book, Algebra: Chapter 0.

I am currently focused on Chapter I, Section 5: Universal Properties.

I need some help Example 5.3 and Exercise 5.2 in this section.
QUESTION 1Example 5.3 reads as follows:View attachment 4484

Can someone clearly explain why $$\emptyset$$ is initial ...

We need every set $$A$$ in the category Set to have exactly one morphism (set function):

$$\emptyset \rightarrow A$$

and Aluffi seems to be saying that the 'empty graph' ( an "empty or nothing function" ? ) is the unique function from $$\emptyset$$ to $$A$$ ... is that right ...?

It seems a highly contrived function (morphism) ... indeed, what is the empty graph exactly ... and how do we rigorously justify the assertion that it is actually a function ...

Can someone please explain clearly just what Aluffi is saying here ... ?
QUESTION 2

Exercise 5.2 in Chapter I, Section 5 reads as follows:View attachment 4485Can someone show me how to construct a rigorous and formal proof for this exercise ... ?Help will be much appreciated ...

PeterPeter

 

[Fallen Angel]

Hi Peter,

It could be useful to remember that a function between two given sets $A, B$ is a subset of $A \times B$ (with some other properties, of course)

 

[Euge]

Peter,

I've given this explanation to you before (that $\emptyset$ is initial in Set), but it's been a while and it doesn't hurt to reiterate. ;) Let $A$ be a set. A morphism $\emptyset \to A$ is a relation from $\emptyset$ to $A$, i.e., a subset $R$ of $\emptyset \times A$ such that for every $x \in \emptyset$, there is a unique $a \in A$ such that $(x,a) \in R$. Now since $\emptyset \times A = \emptyset$, $R = \emptyset$. So there is only one morphism $\emptyset \to A$.

Now let $X$ be a set such that for every set $A$, there is exactly one morphism $X \to A$. Then in particular, there is only one moprhism $X \to \emptyset$. Every element of $X$ maps to only one element of $\emptyset$; this cannot happen unless $X = \emptyset$. There, $\emptyset$ is the unique initial object in Set.

 

[Math Amateur]

Peter,

I've given this explanation to you before (that $\emptyset$ is initial in Set), but it's been a while and it doesn't hurt to reiterate. ;) Let $A$ be a set. A morphism $\emptyset \to A$ is a relation from $\emptyset$ to $A$, i.e., a subset $R$ of $\emptyset \times A$ such that for every $x \in \emptyset$, there is a unique $a \in A$ such that $(x,a) \in R$. Now since $\emptyset \times A = \emptyset$, $R = \emptyset$. So there is only one morphism $\emptyset \to A$.

Now let $X$ be a set such that for every set $A$, there is exactly one morphism $X \to A$. Then in particular, there is only one moprhism $X \to \emptyset$. Every element of $X$ maps to only one element of $\emptyset$; this cannot happen unless $X = \emptyset$. There, $\emptyset$ is the unique initial object in Set.

Thanks for for that post, Euge ... It is most helpful and clear ...

My apologies for forgetting the advice and help that you gave some time earlier ...

Peter

- - - Updated - - -

Hi Peter,

It could be useful to remember that a function between two given sets $A, B$ is a subset of $A \times B$ (with some other properties, of course)

indeed ... a helpful point that is essentially the clue to the situation ... Thanks Fallen Angel ..

Peter