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Section 4.9 Exact sequences (Optional)

We define an useful notion of exact sequences. Throughout this section we assume that \(X,Y, Z,\) and \(W\) are vector spaces (not necessarily finite-dimensional) over a field \(F\text{.}\)

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.

  1. The trivial subspace \(\{0\}\) of a vector space will be denoted by \(0\) when used in a diagram.

  2. 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.

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*}
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 \(Y\text{.}\)
By Definition 4.9.5, \(f\) is injective, \(g\) is surjective, and \(\Im(f)=\ker(g)\text{.}\) Let \(\{x_1,\ldots,x_r\}\) be a basis for \(X\text{.}\)

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*}
By Rank-Nullity Theorem (see Theorem 4.5.3), applied to the linear transformation \(g\colon Y\to Z\text{,}\) we get that
\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.