I'm reading a text for one of my classes, and I came across this theorem:
Theorem 4. Suppose $V$ is an $m$-dimensional vector space over $\mathbb F_p$. Suppose $T : V → W$ is a linear map. Write $n$ for the dimension of the null space of $T$, and $r$ for the dimension of the range. Then
a) The cardinality of $V$ is $|V | = p^m$.
b) The cardinality of the range of T is $p^r$.
c) The preimage of every vector in the range of $T$ has $p^n$ elements.
I don't understand why (c) is true. I can sort of see how given the inverse of an element in the range (call it $x$), one can get $p^n$ more elements mapping to said element by taking $T(x+u)$ where $u$ is in the null space. But then we'd have $q(1+p^n)$ elements in the preimage of every vector $w \in W$, where $q$ is the number of distinct elements of $(V - \text{null} \,V)$ mapping to $w$.
Could someone give me some intuition as to why (c) is true?
If $T$ is a linear map $V \to W$ then, for every $v\in V$, it is the case that: $$T^{-1}(\{Tv\})=\ker(T)+v.$$
Indeed, if $x \in \ker(T)$, then $T(x+v)=Tx+Tv=Tv$, so $x+v\in T^{-1}(\{Tv\})$. Viceversa, if $v' \in T^{-1}(\{Tv\})$, then $T(v'-v)=Tv'-Tv=Tv-Tv=0$ and $v'=(v'-v)+v\in\ker(T)+v$.
Moreover the map $\ker(T)\ni y \mapsto y+v \in\ker(T)+v=T^{-1}(\{Tv\})$ is obviously a bijection. Hence the preimage of every vector in the range of $T$ has the same cardinality of the kernel.
Hence we are done if we prove that the cardinality of the kernel is $p^n$. But this is true, because it is a linear space of dimension $n$ over a field of cardinality $p$. Indeed, if $\{e_1,...,e_n \}$ is a basis of $\ker(T)$, then every element is written uniquely as $a_1e_1+...+a_ne_n$ for $a_i \in \mathbb{F}_p$. Then the linear application: $$ \ker(T)\ni a_1e_1+...+a_ne_n \mapsto (a_1,...,a_n) \in\mathbb{F}_p^n$$ is bijective and then $|\ker(T)|=|\mathbb{F}_p^n|=p^n$.