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Consider the exact sequence $$ \cdots \rightarrow 0 \rightarrow \mathbb{Z} \overset{\times p}{\rightarrow} \color{blue} {\mathbb{Z}} \twoheadrightarrow \mathbb{Z} /p \mathbb{Z} \rightarrow 0 \rightarrow \cdots$$ Show that in the middle group $\color{blue} {\mathbb{Z}}$, the closed elements (or co-cycles) are the elements $p \mathbb Z$ and these are clearly the exact elements in this group.

My approach:

Multiplication map by $p$ takes $ x\in \mathbb Z$ to $p \mathbb Z$. Is this enough to say?

José Carlos Santos
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MAS
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  • No it's not enough, because it doesn't show that you know what you're talking about. You need to recall the definitions of closed and exact. That's about the only difficulty in this question, so you can't get around it. – Arnaud Mortier Nov 27 '19 at 08:36
  • @ArnaudMortier, I understood what I am saying. For, multiplication map by $p$ takes $x \in \Bbb Z$ to $px \in p \Bbb Z$ . Next, closed means the elements of Kernels of the morphisms $\sigma$ between the objects while exact are the elements of image of $\sigma$. Now what to say to to conclude the question? – MAS Nov 27 '19 at 10:58
  • I wasn't claiming that you didn't understand, only that it was not made obvious. If you know that you need to describe a kernel and an image, then simply do that. Your attempt right now only describes an image. – Arnaud Mortier Nov 27 '19 at 11:20
  • @ArnaudMortier, ok, thanks – MAS Nov 27 '19 at 11:26

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