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Let $M_i$ be R-modules and $f_i$ be homomorphisms of R-modules

If $\forall _n\ker f_n=\operatorname{im}f_{n-1}$, wouldn't that mean that for $$...\:\rightarrow M_{n-1}\:\rightarrow ^{f_{n-1}}\:M_n\:\rightarrow ^{f_n} \:M_{n+1}\:\rightarrow \:...$$ we'd get that $M_{n+1} = 0$? Because we'd basically have that $$f_n\left(\operatorname{im}f_{n-1}\right)=f_n\left(\ker f_n\right)=0$$ Or do I get something wrong here?

Bernard
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user
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    Consider the map $f\colon\mathbb{Z}\to\mathbb{Z}$ given by $f(a)=3a$, and the map $g\colon \mathbb{Z}\to \mathbb{Z}/3\mathbb{Z}$ given by the canonical projection. Then $0\to \mathbb{Z}\stackrel{f}{\to}\mathbb{Z}\stackrel{g}{\to}\mathbb{Z}/3\mathbb{Z}\to 0$ is exact, but not every module or map is zero. – Arturo Magidin Apr 17 '21 at 22:54
  • What's a canonical projection? – user Apr 17 '21 at 22:56
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    $\pi(a) = a+3\mathbb{Z}$. – Arturo Magidin Apr 17 '21 at 22:58
  • But why should ker g = im f here? I don't quite see it – user Apr 17 '21 at 22:59
  • Because the image of $f$ is exactly the multiples of $3$, and the kernel of $g$ is precisely the $a$s such that $a+3\mathbb{Z}=3\mathbb{Z}$, that is, exactly the multiples of $3$. – Arturo Magidin Apr 17 '21 at 23:00
  • ah ok now i see it, instead of $\mathbb{Z}/3\mathbb{Z}$ we could've also written $\mathbb{Z}_3$. Now i see it, thanks! Also maybe another question: What's the use of exact sequences? Why do we matter to introduce them? – user Apr 17 '21 at 23:12
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    They are very useful, and if you’ve only now seen the introduction, don’t you think you should have a little patience and see how they are deployed? It’s like asking what’s the point of derivatives after only seeing the definition, and not the myriad of applications. – Arturo Magidin Apr 17 '21 at 23:14
  • It's just the way I study, I need to at least kind of know the application of new stuff being introduced. For derivatives you could say they "calculate the slope at any point x" for example, or even something like "they are very usable in Physics especially Mechanics, Dynamics, ..." – user Apr 17 '21 at 23:15
  • They are a compact way of giving you a lot of information. They are useful in studying modules. They are required to define derived functors. (Again: you need to have some patience. If you need motivation to learn patience, let’s put it this way: you don’t have the tools to understand their use, because you need them before you can understand their use; that’s why you should have patience. Some definitions can be motivated; some just have to be accepted and seen in action to be appreciated.) – Arturo Magidin Apr 17 '21 at 23:29
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    The computation in the question correctly shows that $f_n(\text{im }f_{n-1})=0$, which means that $f_n\circ f_{n-1}=0$. But that doesn't mean $M_{n+1}=0$, as the examples in the answers show. There seems to be an unsupported assumption that $f_n\circ f_{n-1}$ is surjective, probably resulting from unsupported assumptions that $f_n$ and $f_{n-1}$ are surjective. – Andreas Blass Apr 17 '21 at 23:48
  • As the answers show, and is easy to see, an exact sequence certainly doesn't have to be zero. But, the homology of a chain complex is zero when it is exact. Not sure if this is what you were thinking. A chain complex is a sequence in which the image of each map is contained in the kernel of the next one. –  Apr 19 '21 at 20:53

5 Answers5

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A simple and not entirely trivial example of an exact sequence is the exact sequence that you get from any homomorphism $f:M\to N$: $$ \ker(f)\xrightarrow{\quad\iota\quad} M\xrightarrow{\quad f\quad}N\xrightarrow{\quad\pi\quad} N/\operatorname{im}(f) $$ Here, $\iota$ is just the inclusion and $\pi$ the canonical projection. Note that $f\circ\iota=0$ because that's simply how the kernel of a morphism is defined. Furthermore, $\pi\circ f=0$ because for every $m\in M$, you have $f(m)\in\operatorname{im}(f)$, so $\pi(f(m))=0$.

I am sure you can come up with $M$, $N$, and $f$ so that not all of them are zero - I hope this helps to clear things up a bit.

Jesko Hüttenhain
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$ 0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z} /n \mathbb{Z} \to 0$, where $\mathbb{Z} \to \mathbb{Z}, 1 \mapsto n$.

Invincible
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For an infinite exact sequence, consider the $\mathbb{Z}$-module $C_2\times C_2$, and the map $f\colon C_2\times C_2\to C_2\times C_2$ given by $f(a,b) = (b,0)$. Then $$\cdots \stackrel{f}{\longrightarrow} C_2\times C_2 \stackrel{f}{\longrightarrow} C_2\times C_2 \stackrel{f}{\longrightarrow} C_2\times C_2 \stackrel{f}{\longrightarrow}\cdots$$ is exact, since $\mathrm{Im}(f) = \mathrm{ker}(f) = C_2\times\{0\}$. Note that none of the modules are the zero module, and none of the maps are the zero map.

Arturo Magidin
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Here is the standard example of an exact sequence, corresponding to a homomorphism $f:M\longrightarrow N$ of $R$-modules: $$0\longrightarrow\ker f\xrightarrow{\;i\enspace} M\xrightarrow{\;f\enspace} N\xrightarrow{\;p\enspace}\operatorname{coker f}\longrightarrow 0$$ where $i$ is the canonical injection and $p$ the canonical surjection.

Bernard
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As a familiar example, why not

$$ 0 \to \mathbb{R} \to C^\infty(\mathbb{R}) \overset{\frac{d}{dx}}{\to} C^\infty(\mathbb{R}) \to 0 $$

This says that the kernel of the differentiation map is the space of constant functions, and every smooth function on $\mathbb{R}$ is the derivative of some function.

I learned this example a while ago this relevant MO post.

I hope this helps ^_^

Chris Grossack
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