'Pointed sets' are not explicitly defined in the book and I have posted some instances of where they are mentioned according to increasing page number.

Chapter 1

Page 19 (As far as I know, this is the first mentioned of pointed set in the book) enter image description here

Page 24 enter image description here

Chapter 2

Page 43 enter image description here

Page 64

enter image description here

I understand the construction in Chapter 1 page 24: Example 3.8, as well as the answer given here why does unique identity make groups pointed sets?, but I feel that the accepted answer does not address the reason as to why the uniqueness of the identity makes 'groups pointed sets' in the sense of Example 3.8 on page 24 (Assuming uniqueness of the identity is relevant at all).


  1. According to https://en.wikipedia.org/wiki/Pointed_set , a pointed set is just a pair $(X,x)$ where $X$ is a set and $x\in X$. But I'm not sure if this is what Aluffi means for otherwise, why does he mention in Chapter 2 page 43 right after proving that the identity is unique in any arbitrary group that this 'makes groups pointed sets' in the sense of Example 3.8 on page 24? So to me it seems that the author is suggesting that the uniqueness of the identity is a contributing factor to groups being pointed sets? But on the other hand considering Example 3.8 does not suggest anything about the requirement of uniqueness in the context of groups. So is uniqueness of the identity in groups important in establishing that groups are pointed sets?
  1. https://ncatlab.org/nlab/show/pointed+object defines pointed object $X$ to be an object equipped with a global element $1\to X$ where a global element is just a morphism from a terminal object $1$ to $X$. A pointed set is the defined to be a pointed object is $\mathbf{Set}$. Now if I take this definition of a pointed set then every non-empty set in $\mathbf{Set}$ is a pointed object, which isn't quite what the author had in mind when compared to Example 3.8? Or maybe I didn't understand the definition given on nLab. Is it possible to show via example that this definition in nLab is indeed equivalent?
  1. To complicate things further, if I use the definition in nLab then in Chapter 2 page 64 $\text{Hom}_{\mathbf{Grp}}(G,H)$ being a pointed set does not make sense to me since I don't even know what category this is in as an object. So what does Aluffi mean by 'pointed set'? What made $\text{Hom}_{\mathbf{Grp}}(G,H)$ a pointed set? Is it merely that it's not empty, or is there something 'special' about the trivial morphism that makes $\text{Hom}_{\mathbf{Grp}}(G,H)$ into a pointed set? At this point, I don't even know what pointed set means anymore and I feel perplexed at this point.
  • 3
    $\begingroup$ The text on page 24 actually does give a definition: it defines what an object in $\mathsf{Set}^*$ is, and then later it tells you that "pointed set" means an object of $\mathsf{Set}^*$. $\endgroup$
    – N. Virgo
    Sep 8, 2021 at 11:22
  • 1
    $\begingroup$ A pointed set is simply a set containing (among potentially other elements) a distinguished element, called a point. $\endgroup$ Sep 8, 2021 at 11:23
  • 2
    $\begingroup$ To address your question 2: a pointed set is not just a set $A$ such that there exists a morphism $1\to A$, but rather it's the set together with a particular choice of morphism $1\to A$. This is equivalent to a set together with a choice of distinguished element. So $(\{1,2,3\},1)$ and $(\{1,2,3\},2)$ are two different pointed sets. $\endgroup$
    – N. Virgo
    Sep 8, 2021 at 11:26
  • $\begingroup$ What's the stipulation on a distinguished element? I'm guessing it means different things in different contexts? For example, in an arbitrary group it happens to be the identity and in $\text{Hom}_C(G,H)$ it is the trivial morphism. But even if this is the case I still don't see how this definition of pointed set coincides with the one given in nLab. $\endgroup$
    – tcmtan
    Sep 8, 2021 at 11:50
  • 1
    $\begingroup$ @Nathaniel: Thanks for commenting on the choice morphism. It makes sense to me now :) $\endgroup$
    – tcmtan
    Sep 8, 2021 at 23:25

1 Answer 1


In general set theory, a "set" is a collection of objects with no distinction between them and the "morphisms" are any mapping from one set to another.

A "pointed set" is a collection of objects with a single object chosen as the special "point". Morphisms are only those mappings from one "pointed set" to another that map the "special point" of one set to the "special point" of another.

For example, let $X= \{a, b, c\}$ and $Y= \{x, y, z\}$. Considered as sets, we can have $3 \times 3\times 3= 27$ different mappings from $X$ to $Y$.

As "pointed sets" we might have $X= \{a, b, c\mid a\}$, with $a$ selected as the "special point" and $Y= \{x, y, z \mid y\}$ with $y$ selected as the "special point". Now any mapping must map $a$ to $y$ so we have only $3 \times 3= 9$ different mappings from $X$ to $Y$.

  • 2
    $\begingroup$ Crystal clear answer. $\endgroup$
    – Randall
    Sep 8, 2021 at 16:09
  • $\begingroup$ Thank you for this answer, it clarified my understanding. In view of my my first question, it therefore seems to me that the in the context of groups that uniqueness of identities in groups is irrelevant then? What matters is that if we're in $\mathbf{Grp}$ then since group homomorphisms send identities to identities then this guarantees commutativity of the diagram via a group homomorphism (in fact any group homomorphism since our special point is the identity) between the pointed sets? $\endgroup$
    – tcmtan
    Sep 8, 2021 at 23:24
  • $\begingroup$ In view of question 3, I presume that there is some category where $\text{Hom}_{\mathbf{C}}(G,h)$ with some morphism $1\to \text{Hom}_{\mathbf{C}}(G,h)$ (where your distinguished element is the trivial morphism) such that $1$ is a terminal object where this makes sense? Whatever that category is... $\endgroup$
    – tcmtan
    Sep 8, 2021 at 23:30

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