Show that If a sequence is exact the dual sequence is exact

Question

The book is Bosch Linear Algebra $p.79$

We look at at this sequence:

$U\overset{f}\rightarrow V\overset{g}\rightarrow W$

where $f$ and $g$ is linear.

The respective dual sequence is:

$W^*\overset{g^*}{\rightarrow}V^*\overset{f^*}\rightarrow U^*$

where if $V$ is a vectorspace, $V^*$ is the dualspace defined as the set of $\text{Hom}(V,K)$, i.e all linear maps from $V$ to its field which we call $K$. For a linear map $f:U\rightarrow W$, $f^*$ is defined as $f^*:W^*\rightarrow U^*$, with $f^*(\varphi)=\varphi\circ f$, this map is also linear.

A sequence is exact if and only if $\text{im}f=\ker g$

I have to show if a sequence is exact the resp. dual sequence is exact too.

I don't understand the proof, in particular how the Fundamental theorem on homomorphisms is used here, which states if $f:V\rightarrow W$ is a homomorphism between two vectorspaces and there exists a linear subset $U$ of $V$ such that $U\subset \ker f$ then there exists (exactly one) $\bar{f}:V/U\rightarrow W$ such that $f=\bar{f}\circ\pi$ where $\pi$ is the canonical epimorphism

________________$

I have to prove that $\ker f^*=\text{im } g^*$

I already know $\text{im }g^*\subset \ker f^*$ and if $\varphi\in \ker f^{*}$, ie a linear form $\varphi:V\rightarrow K$ with $\varphi\circ f=0$ then $\ker g\subset \ker \varphi$

I understood to show that $\varphi\in \text{im } g^*$ we have to construct a $\psi\in W^*$ such that $\varphi=\psi\circ g$.

We start with splitting $g$ in $g_1$ and $g_2$ where $g_1$ is the surjective linear map to $\text{im }g$ and $g_2$ is the nesting in $W$.

Here is the part I don't understand:

Applying the Fundamental theorem on homomorphisms there exists a $\varphi':\text{im }g\rightarrow K$ such that $\varphi=\varphi'\circ g_1$

We know that $\varphi$ is a linear map. And we also know that $\ker g$ is a subspace and a subset of $\ker\varphi$ then there exist a $\bar{\varphi}:V/\ker g\rightarrow K$ such that $\varphi=\bar{\varphi}\circ \pi$. But the domain is still $V$ and we are looking for a map $\varphi'$ with the domain $\text{im } g$ and the codomain $K$ such that $\varphi=\varphi'\circ g_1$. So this cannot be the map we are looking for. I am not sure if we even can use this map we have constructed but that seemed to me the only reasonable application of the homomorphism Theorem. If I am not wrong and we actually have to use this map, then tell me how I should continue.

Answer 1

$\newcommand{\im}{\operatorname{im}}$Suppose that $\varphi \in V^*$ is in the kernel of $f^*$. Since $f^*\varphi = \varphi \circ f = 0$, we know that $\im f \subseteq \ker \varphi$. Since the original sequence is exact, $\im f = \ker g$. So by the Fundamental Theorem on Homomorphisms (applied to $\varphi$), there is a unique map $\tilde\varphi \colon V/\ker g\to K$ such that $\tilde\varphi\circ \pi = \varphi$, where $\pi$ is the canonical epimorphism from $V$ to $V/\ker g$.

By the same theorem applied to $g_1$, there is a unique map (in fact, an isomorphism) $\tilde g_1 \colon V/\ker g \to \im g$ such that $\tilde g_1 \circ \pi = g_1$. Let $\varphi' = \tilde\varphi \circ \tilde g_1 ^{-1} \colon \im g \to K$. Then $$ \varphi' \circ g_1 = \tilde\varphi\circ\tilde g_1^{-1} \circ g_1 = \tilde\varphi\circ\pi = \varphi $$ as desired.

For the rest of the construction, you need to extend $\varphi'$ to be defined on all of $W$ instead of just $\im g$. There isn't a canonical way to do this without extra structure on $W$, but there are many choices.

Answer 2

Came up with a possible solution, request a short glance over it, because I made a lot of assumptions that were not proved in the book as far as I can remember. Like if two vectorspaces have the same Dimension then the immage of a Basis is alsoa Basis in the other vectorspace.

We look at a complement $Z$ of $\ker g$ and define their bases $A,B$ resp then

$\pi(v)=\pi(\sum_{j\in J\subset A}\lambda_j j+\sum_{i\in I\subset B}\lambda'_i i)=\sum_{j\in J\subset A}\lambda_j j+\ker g$

If $A$ is a basis of $Z$ then $g(A)$ is a Basis of $\text{im }g$. And there exists an isomorphic map $\phi$ between $g(A)$ and $A$

All in all we can say

$\pi(\phi(g_1(v)))=\pi(\phi(g_1(v)))=\pi(\phi(g_1(\sum_{j\in J\subset A}\lambda_j j+\sum_{i\in I\subset B}\lambda'_i i)))=\pi(\phi(g_1(\sum_{j\in J\subset A}\lambda_j j)))$

because $g_1(v)=0$ if $v$ is in the kernel and

$\pi(\phi(g_1(\sum_{j\in J\subset A}\lambda_j j)))=\pi(\sum_{j\in J\subset A}\lambda_j j)=\sum_{j\in J\subset A}\lambda_j j+\ker g$

Therefore the desired map $\varphi'$ is $\pi\circ \phi$

Do you tthink this is Right?