# I Categories of Pointed Sets - Aluffi, Example 3.8

1. May 3, 2016

### Math Amateur

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

I am currently focussed on Section I.3 Categories ... ... and am trying to understand Example 3.8 which is introduced as a concrete instance of the coslice categories referred to in Example 3.7 ...

Examples 3.7 and 3.8 read as follows:

Since I do not have a basic understanding of the category of Example 3.8 my questions may not be well formulated ... for which I apologise in advance ...

My questions are as follows:

Question 1

In the above text by Aluffi we read the following:

" ... ... An object in SET* is then a morphism $f \ : \ \{ \ast \} \longrightarrow S$ in Set where $S$ is any set. ... ... "

My question is as follows: what exactly is $\ast$ ... ?

and ... ... is there only one $\ast$ for the category ... or one for each set ... if it is just a singleton for each set why not refer to it as a special element $s \in S$ ...

Question 2

In the above text by Aluffi we read the following:

" ... ... Thus we may denote object of Set* as pairs $(S,s)$ where $S$ is any set and $s \in S$ is any element of $S$ ... ... "

My question is as follows: is there only one special element of $S$ in the category ... ... or are elements $(S, s_1)$ and $(S, s_2)$ in the category Set* where $s_1$ and $s_2$, like $s$, belong to the set $S$.

Question 3

In the above text by Aluffi we read the following:

" ... ... A morphism between two such objects $(S,s) \longrightarrow (T,t)$, corresponds then to a set-function $\sigma \ : \ S \longrightarrow T$ such that $\sigma (s) = t$. ... ... "

My question is as follows: the prescription $\sigma \ : \ S \longrightarrow T$ such that $\sigma (s) = t$ does not tell us how the other elements of $S$ are mapped ... ... ? ... and there are many alternatives ... and hence presumably, many $\sigma$s ... ... ??? ... ... can someone clarify this matter ...

Hope someone can help ...

Peter

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2. May 3, 2016

### Staff: Mentor

Hello Peter!

$\ast$ is the one single element of the set $Ast = \{ \ast \}$. Clearly this set $Ast$ is an object of the category $C = Set$.
$\ast$ hasn't any further role than being fixed. It's not a placeholder for something arbitrary. If you like choose another symbol. As long as it is the only element of a set, it'll do. He could as well have chosen $1 \in \mathbb{N}$ and built $\{ 1 \}$ but then we would discuss why $1$ and not $2$ or $\pi$. It simply has to be a one element set and a fixed one.

Do not mix things up. There aren't any elements in a category. These are called objects in this case.
Since we consider the category of sets, elements may occur as part of an object, which is a set.

Now we have built the category $Set^*$. It contains all morphisms, i.e. functions $f: Ast → S$ as objects. Since these need to be well-defined we can have only one point, element resp., $s_f \in S$ as image: $f(*) = s_f$. Varying $S$ means different objects in $Set^*$, varying $s_f \in S$ also changes to another object in $Set^*$.

"Or are elements" what? $(S, s_1)$ and $(S, s_2)$ are both objects in $Set^*$. If $s_1≠s_2$ they are different objects since
$f(*) = s_1$ and $g(*) = s_2$ define different morphisms in $C = Set$ which build here the objects of $Set^*$.
Confusing? Yes, in the beginning very likely. But this only displays that on such a theoretical level as category theory we must be especially careful with our wording because objects in one category can be morphisms in another.

What do you mean by other elements of $S$?
Since an object $obj_f = f \in Set^*$ or even more precise $obj_f = f \in Objects(Set^*)$ is a morphism $f: Ast → S$ with $f(*) = s_f \in S$, the author just changes notations. $f$ can as well be written by $(S,s_f)$ or $(S,s)$. Please pay attention to the fact that there is no loss of information. Even more: $f$ is totally determined by $s_f = s$.

Now that we know the objects of $Set^*$ we next need to determine the morphisms of the category $Set^*$.
Since morphisms $μ$ are mappings between objects (of here $Set^*$) we need to define $μ: Objects(Set^*) → Objects(Set^*)$.
Because all objects are already determined by an element, i.e. a point resp., $s_f=s$ of an object $S \in C$, it is sufficient to say what $μ$ shall do on those $s$. Therefore $σ(s)=t$ defines such a (one) morphism. Other $s$ or $t$ yield other morphisms.

More explicit: Let $σ: Morphisms(C) → Morphisms(C)$ be defined as $σ(f) = g$, where $f, g \in Morphisms(C) = \{ Objects(C) → Objects(C) \}$ be defined as $f : Ast = \{*\} → S$ with $f(*) = s_f = s$ and $g : Ast = \{*\} → T$ with $g(*) = t_g = t$.
Here $S,T$ are objects of $C$ which are sets. $s,t$ are elements in these sets, resp. Because it is always only one element per set, it gives rise to the adjective $pointed$.
However, instead of $σ(f) = g$ we may as well write $σ(s) = t$ as explained above.
Therefore $σ$ defines a morphism of the category $Set^*$ (which are mappings between morphisms of the category $C$ of sets which are the objects of $Set$.)

And now I'm dizzy.

Last edited: May 3, 2016
3. May 3, 2016

### Math Amateur

Thanks fresh_42 ... but ... Wow ... still reflecting on that ...

My goodness ... it is abstract all right ... I definitely needed some help ...

Now going mad trying to draw diagrams of objects and morphisms ...

Peter

4. May 3, 2016

### Staff: Mentor

You're welcome. It is not that complicated as it sounds (reads). Just keep different things apart and (or) make an example with small sets.
The difficulty comes in because the author uses $s$ for several roles:
1. $s$ is an element of a set $S$ where $S$ is an object of $C = Set$.
2. $s$ is the image of a morphism $f$ of $C$.
3. $(S,s)$ abbreviates this morphism $f: \{*\} → S$ with $f(*) = s$ since to define $f$ you have to specify $S$ and $s \in S$.
4. $σ(s)=t$ should actually be $σ(f)=g$, but it's shorter to concentrate the following: $"σ(f)$ where $f(*)=s \in S$ and $g(*) = t \in T$ by simply writing $"σ(s) = t "$.

The concept $Set^*$ is a simple example for the general concept of duality.
Perhaps you have met dual vector spaces $V^*$ which consist of all linear function from $V → ℝ$ (for real vector spaces). Here the objects of consideration are mappings, too.

5. May 4, 2016

### Math Amateur

fresh_42 ... Well ... your clarification was REALLY helpful .... THANKS ...

Peter