0

Let $A=B=C=\{0\}$.

Is $A\stackrel{f}\to B\stackrel{g}\to C$ a splitting exact sequence?

We have $Im(f)=ker(g)$ and there is an $h:C\to B$ such that $g\circ h=id_C$ so it should yield a splitting exact sequence but it's so trivial I don't know whether this is true.

Buh
  • 1,395
  • 7
  • 10
  • "it's so trivial I don't know whether this is true" Generally, the solution to this is to read the relevant definition carefully word-for-word, because (usually) it either explicitly excludes your trivial example, or it doesn't, in which case your example is allowed. – Arthur Dec 04 '17 at 13:42
  • That's why I'm here. In the definition I was given it wasn't excluded. – Buh Dec 04 '17 at 13:43
  • Don't you trust ${0}={0}\oplus{0}$? – egreg Dec 04 '17 at 13:46
  • In that case, it is allowed. Usually, when trivial cases are excluded, it is because it leads to cumbersome phrasing down the line ($1$ isn't a prime, because if $1$ were a prime, then almost every single theorem relevant to prime numbers would say "prime other than $1$" all the time). I see no immediate problem with allowing $0\to 0\to 0$ to be called a split sequence. – Arthur Dec 04 '17 at 13:46
  • @Arthur Yes, that is exactly my reason for this question. I remembered that fields are required (for some mysterious reason) to be non-trivial. – Buh Dec 04 '17 at 13:50
  • @Buh Yes, fields are usually required to be non-trivial because the field with one element is a very strange creature. So every result would have "a field with at least $2$ elements" or "a field with $1\neq 0$" to avoid that case. – Arthur Dec 04 '17 at 13:54

1 Answers1

0

Any short exact sequence $0\to A\to B\to C\to 0$ is split whenever $A=\{0\}$ or $C=\{0\}$. The reason is that the definition applies.

egreg
  • 238,574