Note that conditions (1) and (2) in the definition could be replaced by the single condition that if $\vec v_1,\vec v_2\in V$ and $r,s\in\R$, then $f(r\cdot\vec v_1+s\cdot\vec v_2) = r\cdot f(\vec v_1)+s\cdot f(\vec v_2)$. Also note that on the left-hand side of this equation, the scalar multiplication and vector addition are operations in $V$, while on the right-hand side, they are operations in $W$. The same symbols are used for the operations, but it is important not to be confused by that.

The textbook covers isomorphisms first, in Chapter Three, Section I, and it introduces general homomorphisms in Chapter Three, Section II.

One of the major results about isomorphisms is that two finite dimensional vector spaces $V$ and $W$ are isomorphic if and only if they have the same dimension. If $B=\langle\vec v_1,\vec v_2,\dots,\vec v_n\rangle$ is a basis of $V$, and $f\colon V\to W$ is an isomorphism, then it can be shown that $\langle\vec f(v_1),f(\vec v_2),\dots,f(\vec v_n)\rangle$ is a basis of $W$, which means that $W$ also has dimension $n$. Conversely, if $V$ and $W$ both have dimension $n$, then let $B=\langle\vec v_1,\vec v_2,\dots,\vec v_n\rangle$ be a basis of $V$ and $D=\langle\vec w_1,\vec w_2,\dots,\vec w_n\rangle$ be a basis of $W$. Any $\vec v\in V$ can be written in a unique way as a linear combination of elements of $B$: $\vec v = c_1\vec v_1+c_2\vec v_2+\cdots+c_n\vec v_n$, and an isomorphism $f\colon V\to W$ can be defined by $$f(c_1\vec v_1+c_2\vec v_2+\cdots+c_n\vec v_n) = c_1\vec w_1+c_2\vec w_2+\cdots+c_n\vec w_n$$ It is easy to see that $f$ is linear and has inverse function $$f^{-1}(c_1\vec w_1+c_2\vec w_2+\cdots+c_n\vec w_n) = c_1\vec v_1+c_2\vec v_2+\cdots+c_n\vec v_n$$ so it is an isomorphism.

Note that this means that every finite-dimensional vector space $V$ is isomorphic to $\R^n$, where $n$ is the dimension of $V$. Also note that $\R^n$ cannot be isomorphic to $\R^m$ when $n\ne m$, since the dimensions of $\R^n$ and $\R^m$ would be different. So, every finite-dimensional vector space is isomorphic to one and only one $\R^n$.

Turing to the case of general homomorphisms, we find a similar situation with regard to bases: If $f\colon V\to W$ is a homomorphism and $B=\langle\vec v_1,\vec v_2,\dots,\vec v_n\rangle$ is a basis of $V$, then $f$ is completely determined by its values, $f(\vec v_i)$ for $i=1,2,\dots,n$, on the basis vectors. This is true since for an arbitrary $\vec v \in V$, it is possible to write $\vec v$ uniquely as a linear combination $\vec v = c_1\vec v_1+c_2\vec v_2+\cdots+c_n\vec v_n$, and then $f(\vec v) = f(c_1\vec v_1+c_2\vec v_2+\cdots+c_n\vec v_n)= c_1f(\vec v_1)+c_2f(\vec v_2)+\cdots+c_nf(\vec v_n)$. Furthermore, if $\vec w_1, \vec w_2, \dots, \vec w_n$ are any $n$ elements of $W$, then it is possible to define a homomorphism $g\colon V\to W$ by defining $g(\vec v_1)=\vec w_1, g(\vec v_2)=\vec w_2,\dots, g(\vec v_n)=\vec w_n$ and then extending the definition to other vectors in $V$.

Let $f\colon V\to W$ be any homomorphism. Then $f(\vec 0_V) = \vec 0_W$. That is, when $f$ is applied to the zero vector in $V$, the output is the zero vector in $W$. (I have written $\vec 0_V$ and $\vec 0_W$ to remind you that the two zero vectors are in two different spaces, but usually we would write them both as $\vec 0$ or in whatever other form is appropriate for the particular vector spaces in question.) This is true since $f(\vec 0_V) = f(0\cdot \vec 0_V) = 0\cdot f(\vec 0_V) = \vec 0_W$.

Another important fact is that certain sets associated with a linear function are in fact subspaces. Suppose $f\colon V \to W$ is a homomorphism. The set $\{\vec v\in V\,|\,f(\vec v)=\vec 0\}$ is a subspace of $V$ known as the null space or kernel of $f$. The textbook denotes the null space of $f$ as $\mathscr N(f)$. Note that $\mathscr N(f)$ is a subspace of the domain, $V$

The set $\{f(\vec v)\,|\, \vec v\in V\}$ is also a subspace, known as the range space or image of $f$. The textbook denotes the range space as $\mathscr R(f)$. Note that $\mathscr R(f)$ is a subspace of the codomain, $W$.

For a homomorphism $f\colon V\to W$, we define the rank of $f$ to be the dimension of the range space, $\mathscr R(f)$. And we defined the nullity to be the dimension of the null space, $\mathscr N(f)$. The rank of a homomorphism is related to the rank of a matrix, but we won't understand that until we understand how matrices are related to homomorphisms. We have the important theorem