This might seem a strange way of doing things, that is, inventing a possible example (according to comments, there is no such thing as an exact sequence of sets), but let us try to make one for exemplary purposes. The questions are at the bottom.


As noted in a comment, there are no short exact sequence of sets and so I changed this to a long exact sequence of sets as we don't have to worry about the initial and final $\rightarrow$'s, only a finite set of exact intermediate terms.

We let all the morphisms $f$ (i.e. $\alpha,\beta$ etc.) acting on any set $S$, by acting element wise and for $x \in S$ if $x \in im(f)$ then $f(x) = x$ and $0$ otherwise.

\begin{array} \mathrm{\cdots } &\longrightarrow & \{B,C,D,F\} & \overset{i_1} \longrightarrow & \{B,C,D,E,F,H,O,N\} & \overset{\pi_1} \longrightarrow& \{E,H,O,N\} &\longrightarrow&\cdots \\ & & \downarrow & &\downarrow & & \downarrow & &&\\ \mathrm{\cdots} &\longrightarrow & \{A,B,D,M\} & \overset{i_2} \longrightarrow & \{A,B,D,E,H,M,N,O\} & \overset{\pi_2} \longrightarrow & \{E,H,O,N\} &\longrightarrow &\cdots \end{array} where

$\alpha:\{B,C,D,F\} \rightarrow \{A,B,D,M\}$

$\eta: \{B,C,D,E,F,H,O,N\} \rightarrow \{A,B,D,E,H,M,N,O\}$

$\beta: \{E,H,O,N\} \rightarrow \{E,H,O,N\}$

Now to show the concept of a pushout, that if another set say $\{A,B,D,K,O,H\}$ in which:

$\alpha ':\{A,B,D,M\} \rightarrow \{A,B,D,K,O,H\}$ and

$\eta ':\{B,C,D,E,F,H,O,N\} \rightarrow \{A,B,D,K,O,H\}$

$\rho :\{A,B,D,E,H,M,N,O\} \rightarrow \{A,B,D,K,O,H\}$

We find that $\eta \circ i_1 = \{B,D\} = \alpha \circ i_2$ But also checking $\eta ' \circ i_1 =\{B,D\} = \alpha ' \circ \alpha$

Looking at the diagram from here commutative diagram for a pushout Here is the diagram I am interested in: commutative diagram of interest

Then $\{A,B,D,E,H,M,N,O\}$ should be the pushout!

The questions are:

1.I observe: $(\{A,B,D,M\} \cup \{B,C,D,E,F,H,O,N\}) mod \{C,F\} = \{A,B,D,E,H,M,N,O\}$ That this has something to do with: $\{A,B,D,E,H,M,N,O\} = (\{A,B,D,M\} \oplus \{B,C,D,E,F,H,O,N\})\, mod \, T$ where $T= \{(-\alpha(x),i_1(x)) \in (\{A,B,D,M\} \oplus \{B,C,D,E,F,H,O,N\}) \forall x \in \{B,C,D,F \}$. I don't see how this relates to the notation such, especially $-\alpha(x)$, what are they trying to say?

  1. While I see that $\{A,B,D,E,H,M,N,O\} = \{A,B,D,M\} \oplus \{E,H,O,N\}$ and $\{A,B,D,E,H,M,N,O\}$ is an extension of $\{E,H,O,N\}$ by $\{A,B,D,M\}$ How would I prove this in a general case?

    1. if $\{A,B,D,E,H,M,N,O\}$ is the pushout of $i_1$ and $\alpha$ then is $\{E,H,O,N\}$ the pushout of $i_2$ and $\eta$?



  • 2
    $\begingroup$ I don't know much about this but I think you can't speak of exact sequences (or at least not like this) for sets. The problem is that unlike modules, sets do not have a $0$ object, because the initial object $\emptyset$ and the final object $\{*\}$ are not isomorphic. In fact if $X$ is non-empty there is no map $X\to \emptyset$. As far as I know, if you want to speak of exact sequences of sets you need to use kernel pairs. $\endgroup$ – Arnaud D. Sep 2 '15 at 16:33
  • $\begingroup$ That's right-there's no such thing as an exact sequence of sets. $\endgroup$ – Kevin Carlson Sep 2 '15 at 17:02
  • $\begingroup$ Extremely useful Arnaud! Thanks - actually the elements in the sets I have above I use in my work to denote particular partitions of a 4 atom Venn diagram. What you mentioned will definitely come in handy in the future. I believe I see why $\emptyset$ and $\{*\}$ are not isomorphic, that being that the former doesn't have an element and the latter does. Please let me know if that is the reasoning. $\endgroup$ – Relative0 Sep 2 '15 at 18:50
  • $\begingroup$ @ArnaudD. There is such a thing as an "exact sequence" of pointed sets, though: $A \to B \to C$ is exact if the preimage of the base point of $C$ is equal to the image of $A \to B$. It appears e.g. in Bousfield homotopy spectral sequences. It doesn't appear to be the case here though. $\endgroup$ – Najib Idrissi Sep 3 '15 at 7:45
  • $\begingroup$ Very useful Najib! Further on down the road I will have a base point (pun intended) to see how at least pointed sets are related in exact sequences. $\endgroup$ – Relative0 Sep 3 '15 at 15:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.