MHB Categories - Bland Chapter 3 - Problem 2 - Problem Set 3.1

[Math Amateur]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am trying to understand Section 3.1 on Categories.

At present I am working on Problem 2 in Problem Set 3.1 and I need some help in understanding the problem and its solution.

Problem 2 (Problem Set 3.1) reads as follows:https://www.physicsforums.com/attachments/3602Now in the category Set we have:

$$\mathscr{O}$$ is the class of all sets - that is the objects are sets

and

If $$A, B \in \mathscr{O} $$ then $$\text{Mor} (A,B)$$ is the set of all functions from $$A$$ to $$B$$ ...

Now Bland's definition of an initial and final object are as follows:https://www.physicsforums.com/attachments/3603

Now if $$\emptyset$$ is an initial object, then $$\text{Mor} (\emptyset, B)$$ has exactly one morphism $$f \ : \ \emptyset \rightarrow B$$ ... ...

... ... BUT ... ... how can we have a set function emanating from a set with no elements ...

Can someone please clarify this issue/problem?

Peter

 

[Fallen Angel]

Hi Peter,

I have never rode such a strange thing, so I'm not really sure of what I'm going to say, but I think here could be an agreement.

If you consider the sets $A,B$ and the set $C=\{f:A\longrightarrow B \ : \ f \ \mbox{is an application}\}$

Then we know that in general $\#(C)=\#(B)^{\#(A)}$ and $\#(\emptyset)=0$, hence we can agree that $C_{\emptyset}=\{f:\emptyset \longrightarrow B \ : \ f \ \mbox{is an application}\}$ has one element.

A different question would be if this agreement make sense or not.

 

[Math Amateur]

Hi Peter,

I have never rode such a strange thing, so I'm not really sure of what I'm going to say, but I think here could be an agreement.

If you consider the sets $A,B$ and the set $C=\{f:A\longrightarrow B \ : \ f \ \mbox{is an application}\}$

Then we know that in general $\#(C)=\#(B)^{\#(A)}$ and $\#(\emptyset)=0$, hence we can agree that $C_{\emptyset}=\{f:\emptyset \longrightarrow B \ : \ f \ \mbox{is an application}\}$ has one element.

A different question would be if this agreement make sense or not.

Thanks for the help, Fallen Angel ...

Hmmm ... not sure ... still reflecting and thinking about this matter ...

Does anyone else have a viewpoint on this ... ... ?

Peter

 

[Euge]

Thanks for the help, Fallen Angel ...

Hmmm ... not sure ... still reflecting and thinking about this matter ...

Does anyone else have a viewpoint on this ... ... ?

Peter

Hi Peter,

You've asked a very good question. The abstract definition of a function $f : A \to B$ is a relation from $A$ to $B$ (i.e., a subset $R$ of $A \times B$) such that for every $a \in A$, there is a unique $b \in B$ such that $(a,b) \in R$ (and we typically write $f(a) = b$ when $(a,b) \in R$). Hence, given an object $B$ of SET and a morphism $f \in \text{Mor}(\emptyset, B)$, $f$ is viewed as a relation from $\emptyset$ to $B$, in other words, a subset of $\emptyset \times B$. Since $\emptyset \times B = \emptyset$, $f$ is determined by the empty relation. Conversely, the empty relation gives a morphism from $\emptyset$ to $B$. Therefore, $\text{Mor}(\emptyset, B)$ has only one element. Since $B$ was an arbitrary object of SET, $\emptyset$ is the initial object of SET.

 

[Math Amateur]

Hi Peter,

You've asked a very good question. The abstract definition of a function $f : A \to B$ is a relation from $A$ to $B$ (i.e., a subset $R$ of $A \times B$) such that for every $a \in A$, there is a unique $b \in B$ such that $(a,b) \in R$ (and we typically write $f(a) = b$ when $(a,b) \in R$). Hence, given an object $B$ of SET and a morphism $f \in \text{Mor}(\emptyset, B)$, $f$ is viewed as a relation from $\emptyset$ to $B$, in other words, a subset of $\emptyset \times B$. Since $\emptyset \times B = \emptyset$, $f$ is determined by the empty relation. Conversely, the empty relation gives a morphism from $\emptyset$ to $B$. Therefore, $\text{Mor}(\emptyset, B)$ has only one element. Since $B$ was an arbitrary object of SET, $\emptyset$ is the initial object of SET.

Thanks for the help, Euge ...

Still reflecting on what you have written, but I get the idea, I think ...

Most helpful ... thanks again ...

Peter