# Preimages of elements in linear maps involving finite fields

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 $$|\ker(T)|=|\mathbb{F}_p^n|=p^n$$.