In linear algebra, we know that the linear transformation $f:V\to W$ induces the linear transformations $V\xrightarrow{g}\text{im}(f)\xrightarrow{i}W$ where $i$ is the inclusion map and $g$ is the same as $f$ except that the range of $f$ has been restricted to $\text{im}(f)$. Then $f=i\circ g$ and so we can represent this by a commutative diagram. Intuitively, we can think of exact sequence as a factorisation of maps with certain condition.
Definition: Let $U\xrightarrow{f}V\xrightarrow{g}W$ be a sequence of linear transformations. (This means $f\in L(U,V)$ and $g\in L(V,W)$.) We say $U\xrightarrow{f}V\xrightarrow{g}W$ is an exact sequence at $V$ if $\text{im}(f)=\text{ker}(g)$. A sequence $V_1\xrightarrow{f_1}V_2\xrightarrow{f_2}\cdots \xrightarrow{f_{n-2}}V_{n-1}\xrightarrow{f_{n-1}}V_n$ is an exact sequence if $\text{im}(f_{i-1})=\text{ker}(f_i), 2\leq i\leq n-1$, that is, if the sequence is exact at $V_2,\cdots,V_{n-1}$.
When writing sequences of linear transformations, it is customary to write $0$ for the vector space $\{0\}$.
Lemma:
1. The sequence $0\to V\to 0$ is exact iff $V=\{0\}$.
2. The sequence $0\to U\xrightarrow{f} V$ is exact iff $f$ is injective.
3. The sequence $V\xrightarrow{g}W\to 0$ is exact iff $g$ is surjective.
4. The sequence $0\to V\xrightarrow{f}W\to 0$ is exact iff $f$ is an isomorphism.
5. The sequence $0\to U\xrightarrow{f}V\xrightarrow{g}W\to 0$ is exact iff $f$ is injective and $g$ is surjective, and $\text{im}(f)=\text{ker}(g)$.
Proof: Just note that from the definition, $\text{im}(0\to V)=0$ and $\text{ker}(V\to 0)=V.\:\Box$
Definition: A short exact sequence is a sequence of five vector spaces, which is exact at all three internal nodes: $0\to U\xrightarrow{f}V\xrightarrow{g}W\to 0$. A long exact sequence is a doubly infinite sequence of vector spaces which is exact at every node.
A nice feature of exact sequences is that there are simple rules to create new sequences from old ones. The simplest way to do this is to split and glue exact sequences as follows.
Splitting and gluing exact sequences
Splitting: If $V_1\xrightarrow{f_1}V_2\xrightarrow{f_2}V_3\xrightarrow{f_3}V_4$ is an exact sequence, the two sequences $V_1\xrightarrow{f_1}V_2\xrightarrow{f_2}W\to 0$ and $0\to W\to V_3\xrightarrow{f_3}V_4$ are also exact, where $W=\text{im}(f_2)=\text{ker}(f_3)$ and $f_2$ is the inclusion of $\text{ker}(f_3)$ in $V_3$.
Gluing: Conversely, if $V_1\xrightarrow{f_1}V_2\xrightarrow{f_2}W\to 0$ and $0\to W\to V_3\xrightarrow{f_3}V_4$ are exact sequences with $W \subset V_3$ and $f_2$ being the inclusion, the sequence $V_1\xrightarrow{f_1}V_2\xrightarrow{f_2}V_3\xrightarrow{f_3}$ is also exact.
Proposition: Assume that $0\to V_1\xrightarrow{f_1}V_2\xrightarrow{f_2}\cdots\xrightarrow{f_{n-1}}V_n\to 0$ is an exact sequence of finite dimensional vector spaces, $n\geq 1$. Then $\sum\limits_{i=1}^n (-1)^i \: \text{dim}(V_i)=0$.
Proof by MI: If $n=1$, then $V_1=\{0\}$, and so $\text{dim}(V_1)=0$.
If $n=2$, then $V_1 \cong V_2$ by the lemma, and so $\text{dim}(V_1)-\text{dim}(V_2)=0$.
If $n=3$, then $0\to V_1\xrightarrow{f_1}V_2\xrightarrow{f_2}V_3\to 0$ is an exact sequence.
By the dimension theorem, we have $$\begin{align}\text{dim}(V_2)&=\text{dim}\:\text{ker}(f_2)+\text{dim}\:\text{im}(f_2)\\
&=\text{dim}\:\text{im}(f_1)+\text{dim}(V_3)\\
&=\text{dim}(V_1)-\text{dim}\:\text{ker}(f_1)+\text{dim}(V_3)\\
&=\text{dim}(V_1)+\text{dim}(V_3),\end{align}$$ which means $-\text{dim}(V_1)+\text{dim}(V_2)-\text{dim}(V_3)=0$. So the result holds for $n=3$.
For the general case, suppose $n\geq 4$. To use the induction assumption, we can split the given exact sequence into $$0\to V_1\xrightarrow{f_1}\cdots\xrightarrow{f_{n-3}}V_{n-2}\xrightarrow{f_{n-2}}\text{im}(f_{n-2})\to 0,$$ $$0\to \text{im}(f_{n-2})\xrightarrow{i}V_{n-1}\xrightarrow{f_{n-1}}V_n\to 0$$ because $\text{im}(f_{n-2})=\text{ker}(f_{n-1})$. By induction, we have $\sum\limits_{i=1}^{n-2} (-1)^i \:\text{dim}(V_i)+(-1)^{n-1}\:\text{dim}(\text{im}(f_{n-2}))=0$ and $(-1)^{n-2}\:\text{dim}(\text{im}(f_{n-2}))+(-1)^{n-1}\:\text{dim}(V_{n-1})+(-1)^n\:\text{dim}(V_n)=0$.
Adding these two gives $\sum\limits_{i=1}^n (-1)^i \:\text{dim}(V_i)=0.\:\Box$
Direct proof: Applying the rank nullity theorem to all vector spaces $V_1,\cdots,V_n$, we have $$\begin{align}\sum\limits_{i=1}^{n-1}(-1)^i\:\text{dim}(V_i)&=\sum\limits_{i=1}^{n-1}(-1)^i \:(\text{dim}\:\text{ker}(f_i)+\text{dim}\:\text{im}(f_i))\\
&=-\text{dim}\:\text{ker}(f_1)+(-1)^{n-1}\text{dim}\:\text{im}(f_{n-1})\\&+\sum\limits_{i=2}^{n-1}(-1)^i \:(\text{dim}\:\text{ker}(f_i)-\text{dim}\:\text{im}(f_{i-1})).\end{align}$$
But $\text{dim}\:\text{ker}(f_1)=0$ since $f_1$ is injective, $\text{dim}\:\text{im}(f_{n-1})=\text{dim}(V_n)$ since $f_{n-1}$ is surjective, and $\text{dim}\:\text{ker}(f_i)=\text{dim}\:\text{im}(f_{n-1})$ for all $i=2,\cdots,n-1$ by definition.
Therefore $\sum\limits_{i=1}^n (-1)^i \: \text{dim}(V_i)=(-1)^{n-1}\:\text{dim}(V_n)+(-1)^n\:\text{dim}(V_n)=0.\:\Box$
Linear algebra
Dimension theorems
For any short exact sequence of finite-dimensional vector spaces $0\to V_1\xrightarrow{T_1}V_2\xrightarrow{T_2}V_3 \to 0$, $\text{dim}(V_2)=\text{dim}(V_1)+\text{dim}(V_3)$. [This is just the $n=3$ case of the proposition above. Here we provide a proof without the aid of rank nullity theorem.]
Proof: Suppose $\text{dim}(V_1)=n$. Then we can take $\{e_i\}_{i=1}^n$ to be a basis of $V_1$. Thus $\{T_1(e_i)\}_{i=1}^n$ span the image of $T_1$. Since $T_1$ is injective, $\{T_1(e_i)\}_{i=1}^n$ are linearly independent in $V_2$. We can find $\{v_j\}_{j=1}^r$ such that $\{T_1(e_i)\}_{i=1}^n \cup \{v_j\}_{j=1}^r$ form a basis of $V_2$. Thus $\text{dim}(V_2)=n+r$. Now $T_2$ is surjective, so $\{T_2(v_j)\}_{j=1}^r$ span $V_3$. It remains to show that $\{T_2(v_j)\}_{j=1}^r$ is linearly independent. Suppose $\sum_j a_jT_2(v_j)=0$. Then $\sum_j T_2(a_jv_j)=0\Rightarrow \sum a_jv_j \in \text{ker}(T_2)=\text{im}(T_1)$. Since $\{T_1(e_i)\}_{i=1}^n$ span the image of $T_1$, we have $\sum\limits_{j=1}^r a_jv_j=\sum\limits_{i=1}^n b_i T_1(e_i)$ for some constants $\{b_i\in \mathbb{R}\}_{i=1}^n$. In particular, $\sum\limits_{i=1}^n -b_i T_1(e_i)+\sum\limits_{j=1}^r a_jv_j=0$. Since $\{T_1(e_i)\}_{i=1}^n \cup \{v_j\}_{j=1}^r$ are linearly independent in $V_2$, all the coefficients $\{b_i\}$ and $\{a_j\}$ must be zero. Therefore, $\text{dim}(V_3)=r$. We have established that $\text{dim}(V_2)=n+r=\text{dim}(V_1)+\text{dim}(V_3).\:\Box$
Using this, we can derive many results in linear algebra.
Theorem: If $T:V\to W$, then $0 \to \text{ker}(T) \xrightarrow{i} V \xrightarrow{T} \text{im}(T) \to 0$ is a short exact sequence, where $i$ is the inclusion map.
Corollary (Rank-nullity): If $T:V\to W$ is a linear map of finite-dimensional vector spaces, then $\text{dim}(\text{ker}(T))+\text{dim}(\text{im}(T))=\text{dim}(V)$.
Theorem: If $A,B \subset V$ are subspaces such that $V=\text{span}(A\cup B)$, then $0\to A \cap B \to A \oplus B \to V \to 0$ is a short exact sequence, where $A \cap B\to A\oplus B$ is the diagonal map $v\mapsto (v,v)$, and the map $A\oplus B\to V$ is the difference map $(a,b)\mapsto a-b$.
Corollary: $\text{dim}(V)=\text{dim}(A)+\text{dim}(B)-\text{dim}(A \cap B)$.
Theorem: If $W\subset V$ is a subspace, then $0\to W \xrightarrow{i} V \xrightarrow{g} V/W \to 0$ is exact, where $i$ is the inclusion map and $g$ is the quotient map $g(v)=[v]$.
Corollary: $\text{dim}(V/W)=\text{dim}(V)-\text{dim}(W)$.
Dual space
Proposition: If $U\xrightarrow{f}V\xrightarrow{g}W$ is an exact sequence of linear transformations of vector spaces, then $W^*\xrightarrow{g^*}V^*\xrightarrow{f^*}U^*$ is also an exact sequence of linear transformations.
Corolloary: Let $f\in L(V,W)$ and let $f^*\in L(W^*,V^*)$ be the induced linear transformation on dual spaces.
1. If $f$ is injective, then $f^*$ is surjective.
2. If $f$ is surjective, then $f^*$ is injective.
Proof:
If $f$ is injective, then $0\to V\to W$ is exact at $V$ and so $W^*\to V^*\to 0$ is an exact sequence. Thus $f^*$ is surjective.
If $f$ is surjective, then $V\to W\to 0$ is exact at $W$ and so $0\to W^*\to V^*$ is an exact sequence. Thus $f^*$ is injective.
More intuition about exact sequences
In Euclidean space $\mathbb{R}^n$, if $V$ is a subspace of $\mathbb{R}^n$, then we think of it as filling out "some of" the dimensions in $\mathbb{R}^n$ and its orthogonal complement $V^\perp$ fills out the other directions. Together they span $\mathbb{R}^n$ with no redundancies: $\mathbb{R}^n=V\oplus V^\perp$. See here for more examples of direct sum.
More generally, we don't have an inner product and so we can't form orthogonal complements, but we can still talk about submodules and quotients.
If $A$ is a submodule of $B$, then $A$ fills up "some of the directions" in $A$, and the remaining directions are encoded in $B/A$. When we have a submodule $A\subset B$, a surjection $B \to C$, and if $A$ happens to be the kernel of the map $B \to C$, then we are in the previous situation: $A$ fills out some of the directions in $B$, and all the complementary directions are encoded in $C$. So we introduce the terminology $0\to A\to B\to C\to 0$ is a short exact sequence to capture this situation.
Graph theory: Euler's formula
Inclusion-exclusion
References
Intuitive meaning of exact sequence
Linear algebra [p38-40,75,76,84]
Exact sequences in linear algebra [p4-6]
Definition of short exact sequence [p1,2]
Splitting and gluing [p1,2,5]
Simple groups and short exact sequences [p2,4]
More to explore:
Homological algebra
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