I'll just quote my book here so you can see the definitions I have:

Suppose that you are given a sequence of vector spaces $V_i$ and linear maps $\varphi_i: V_i\to V_{i+1}$ connecting them, as illustrated below: $$\cdots \longrightarrow V_{i-1} \stackrel{\varphi_{i-1}}{\longrightarrow} V_i \stackrel{\varphi_{i}}{\longrightarrow} V_{i+1} \stackrel{\varphi_{i+1}}{\longrightarrow} \cdots$$ The maps are said to be exact at V_i if $\operatorname{im} \varphi_{i-1} = \operatorname{ker}\varphi_i$. The sequence is called an exact sequence if the maps are exact at $V_i$ for all $i$. $\dots$
If $V_1, V_2$ and $V_3$ are three vector spaces, and if the sequence $$0 \stackrel{\varphi_0}{\longrightarrow} V_{1} \stackrel{\varphi_{1}}{\longrightarrow} V_2 \stackrel{\varphi_{2}}{\longrightarrow} V_{3} \stackrel{\varphi_{3}}{\longrightarrow} 0 \tag{1.7}$$ is exact, it is called a short exact sequence. In this diagram "$0$" represents the zero-dimensional vector space.

OK, here's what I'm not understanding. If the image of any function in this sequence is the kernel of the next function, doesn't every step of this just map to $0$? And even if it didn't, because we're starting with the $0$ vector space, everything has to map to $0$ because linear transformations always map $0$ to $0$. So I'm not understanding this definition at all. The first exercise right below these definitions is to show that equation $(1.7)$ implies that $\varphi_1$ is injective and $\varphi_2$ is surjective. But all I'm seeing here is a chain of functions mapping zero to zero. Can someone explain what I'm missing here?

  • $\begingroup$ All linear transformations map zero to zero anyway. $\endgroup$
    – D_S
    Commented Jun 25, 2016 at 13:35
  • $\begingroup$ Zero always maps to zero -- so you are seeing quite a bit more than a chain of functions mapping zero to zero. The exercise you mentioned at the end is a good one. It isn't particularly hard but doing it will help you to understand the concepts more. $\endgroup$ Commented Jun 25, 2016 at 13:36
  • $\begingroup$ @user1551: how is this related to Clifford algebras? $\endgroup$
    – Watson
    Commented Jun 25, 2016 at 16:24
  • $\begingroup$ @Watson I'll remove the tag. $\endgroup$
    – user1551
    Commented Jun 25, 2016 at 16:51

4 Answers 4


I'll answer with the most important example:

If $T:V \to W$ is a linear transformation between vector spaces, then

$0 \to \text{Nullspace}(T) \to V \to \text{Range}(T) \to 0$

is a short exact sequence, where the map Nullspace$(T) \to V$ is the inclusion, and the map $V \to \text{Range}(T)$ is just $v \mapsto Tv$.

Prove that this is exact.

Once you do, you'll have a whole family of short exact sequences which aren't trivial. Also, all short exact sequences are "isomorphic" to this one for some $T$.

Not really part of the answer, but an important fact about exact sequences (of finite-dimensional vector spaces, and only finitely many of them) is that if you take the alternating sum of dimensions (add all dimensions, but give odd-numbered terms a minus sign), you get zero. In the above example this is equivalent to a familiar fact from linear algebra. What is that fact?


You're right that the composition maps $\varphi_n \circ \cdots \circ \varphi_0$ will always be the zero map. However, this doesn't mean that the individual maps are the zero map. In fact, it's easy to find a pair of nonzero linear maps whose composition is zero. If you follow a single element through from the beginning, you will only get to zero, but if you start somewhere in the middle of the sequence, you can get to other elements.

What the definition of a short exact sequence says is that, for instance, the kernel of $\varphi_1$ must be equal to the image of $\varphi_0$. You know what the image of $\varphi_0$ is, so you therefore know what the kernel of $\varphi_1$ is: 0. That doesn't tell you that the image of $\varphi_1$ is zero; the image of $\varphi_1$ is isomorphic to $V_1/ \ker \varphi_1$. Hopefully this helps.


No. The image of $\varphi_0$, which is $\{0\}$, is the kernel of $\varphi_1$. Thus $\varphi_1$ is injective.

Similarly the kernel of $\varphi_3$, which is $V_3$ is the image of $\varphi_2$, which means $\varphi_2$ is surjective.

We may sum up all this, saying we have a short exact sequence if and only if $\varphi_1$ is one-to-one, $\varphi_2$ is onto and $\ker \varphi_2=\operatorname{Im}\varphi_1$.


Since $\varphi_0 (0)=0$, you have that $\ker\varphi_1=0$, so $\varphi_1$ is injective.

Here is an example. Choose $V $ to be your favorite vector space. Then $$0\to V\to V\oplus V\to V\to0$$ is exact when we take $$\varphi_0=0,\ \ \varphi_1 (v)=v\oplus 0,\ \ \varphi_2 (v\oplus w)=w,\ \ \varphi_3 (v)=0$$ for all $v,w \in V $. Find the kernels and images and you'll see.


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