Definition 4.9.1.
Let \(f\colon X\to Y\) and \(g\colon Y\to Z\) be \(F\)-linear maps. The ordered pair \((f,g)\) is called an exact sequence if
\begin{equation*}
g^{-1}\{0\}=f(X).
\end{equation*}
Section 4.9 Exact sequences (Optional)
Diagram of vector spaces: Let \(f\) be an \(F\)-linear map from \(X\) to \(Y\text{,}\) and \(g\) an \(F\)-linear map from \(Y\) to \(Z\text{,}\) \(h\) be an \(F\)-linear map from \(Z\) to \(W\text{.}\) This information we write schematically in the form of a diagram:
\begin{equation*}
X\xrightarrow{f}Y\xrightarrow{g}Z\xrightarrow{h}W
\end{equation*}
We emphasize that the diagram is not a mathematical object but a valuable tool to facilitate reading an argument.
We make the following conventions.
- The trivial subspace \(\{0\}\) of a vector space will be denoted by \(0\) when used in a diagram.
- Since there is a unique linear map from the trivial subspace to the vector space (resp., from a vector space to its trivial subspace) we will not write the name for this linear map.
Remark 4.9.2.
By Definition 4.4.1 and Definition 4.4.5, the pair \((f,g)\) is an exact sequence if and only if \(\ker(g)=\Im(f)\text{.}\)Definition 4.9.3.
A diagram of vector spaces with \(F\)-linear maps \(f, g\)
\begin{equation*}
X\xrightarrow{f}Y\xrightarrow{g}Z
\end{equation*}
is called an exact sequence if the ordered pair \((f,g)\) is exact, i.e., if \(\ker(g)=\Im(f)\text{.}\)Definition 4.9.4.
Consider the following diagram of vector spaces with \(F\)-linear maps \(f,g\text{,}\) and \(h\text{.}\)
\begin{equation*}
X\xrightarrow{f}Y\xrightarrow{g}Z\xrightarrow{h}W
\end{equation*}
This diagram is said to be exact at Y if \(X\xrightarrow{f}Y\xrightarrow{g}Z\) is exact. Similarly, this diagram is said to be exact at Z if \(Y\xrightarrow{g}Z\xrightarrow{h}W\) is exact.Definition 4.9.5.
A diagram of vector spaces with \(F\)-linear maps \(f,g\)
\begin{equation*}
0\xrightarrow{}X\xrightarrow{f}Y\xrightarrow{g}Z\xrightarrow{}0
\end{equation*}
is called a short exact sequence if \(f\) is injective, \(g\) is surjective, and the diagram is exact at \(Y\text{.}\)Theorem 4.9.6. (Dimension as an additive function).
Let \(X,Y\text{,}\) and \(Z\) be finite-dimensional vector spaces over a field \(F\text{.}\) Consider the following short exact sequence of vector spaces with \(F\)-linear maps \(f,g\text{.}\)
\begin{equation*}
0\to X\xrightarrow{f} Y\xrightarrow{g} Z\to 0
\end{equation*}
We then have
\begin{equation*}
\dim_FX-\dim_FY+\dim_FZ=0.
\end{equation*}
Proof.
We claim that \(\{f(x_1),\ldots,f(x_r)\}\) is a basis for \(\Im(f)\text{.}\) Suppose that \(\sum\alpha_if(x_i)=0\text{,}\) i.e., \(f\big(\sum\alpha_ix_i\big)=0\text{.}\) Hence, \(\sum\alpha_ix_i\in\ker(f)\text{.}\) Since \(f\) is injective, we get \(\sum\alpha_ix_i=0\text{.}\) The linear independence of \(\{x_1,\ldots,x_n\}\) implies that \(\alpha_i=0\) for each \(i\text{.}\) Therefore, \(\{f(x_1),\ldots,f(x_r)\}\) is linearly independent.
We now show that \(\{f(x_1),\ldots,f(x_r)\}\) spans \(\Im(f)\text{.}\) Indeed, suppose that \(f(x)\in\Im(f)\text{.}\) There exists unique \(\alpha_i\in F\) such that \(x=\sum\alpha_ix_i\text{.}\) Hence, \(f(x)=\sum\alpha_if(x_i)\text{.}\)
We thus have proved that \(\{f(x_1),\ldots,f(x_r)\}\) is a basis (this shows that \(X\) is \(F\)-isomorphic to \(f(X)\text{,}\) see Lemma 5.1.4). Therefore, using \(\Im(f)=\ker(g)\text{,}\) we have
\begin{equation*}
\dim_F\ker(g)=\dim_F\Im(f)=r=\dim_FX\text{.}
\end{equation*}
Furthermore, using \(g\) is surjective, we have
\begin{equation*}
\dim_FZ=\dim_F\Im(g).
\end{equation*}
\begin{align*}
\dim_FY\amp=\dim_F\ker(g)+\dim_F\Im(g)\\
\amp=\dim_F\Im(f)+\dim_FZ\\
\amp=\dim_FX+\dim_FZ.
\end{align*}
In other words, \(\dim_FX-\dim_FY+\dim_FZ=0\text{,}\) and hence the theorem is proved.Remark 4.9.7.
In fact following more general result is true.Let \(X_0,X_1,\ldots,X_n\) be finite-dimensional vector spaces over a field \(F\text{.}\) Suppose that we have the following exact sequence with \(F\)-linear maps \(d_i\text{.}\)
\begin{equation*}
0\to X_0\xrightarrow{d_0}X_1\xrightarrow{d_1}X_2\xrightarrow{d_2}\cdots\xrightarrow{d_{n-1}}X_n\to 0
\end{equation*}
Thus, we have \(d_0\) is injective, \(d_{n-1}\) is surjective, and \(\ker(d_{k+1})=\Im(d_{k})\) for each \(0\leq k\leq n-2\text{.}\) Then,
\begin{equation*}
\sum_{i=0}^{n}(-1)^{i}\dim_FX_i=0.
\end{equation*}
We will not prove this result. A possible way to prove this result is by ’splitting’ the exact sequence into several short exact sequences:
\begin{gather*}
0\to X_0\xrightarrow{d_0}X_1\xrightarrow{d_0}\Im(d_0)\to 0\\
0\to\ker(d_k)\xrightarrow{\text{inclusion map}}X_k\xrightarrow{d_k}\Im(d_k)\to 0\quad\text{for each }1\leq k\leq n-1.
\end{gather*}
One may then apply Theorem 4.9.6 for every short exact sequence obtained above to get the result.
