TITLE: What is a short exact sequence?
QUESTION [9 upvotes]: 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?

REPLY [15 votes]: 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?