For a linear function, the fiber of the output is the translate of the kernel by the input. (Trivial observation, proof needed.) As you may already know, I am a newbie to linear algebra.
I am supposed to prove that
for every linear function between vector spaces,
for every input,
the fiber of the corresponding output
equals
the affine subspace which equals
the translate of the kernel by the input.
In other words, I am supposed to prove the following $\dots$

$\textbf{Theorem}$. Let $f$ be a linear function between vector spaces.
For every input $v$,
$$f^{-1}\big( f(v) \big) = \{v + v' \mid v' \in \mathrm{ker} f \}.$$

$\dots$ and I wrote the following proof.
Could you please comment and criticise?
$~$
$\textit{Proof}$
With respect to the above identity,
we denote the left-hand side by
\begin{equation*}
L := f^{-1}\big( f(v) \big).\quad\quad\quad\quad
\end{equation*}
Similarly, we denote the right-hand side by
\begin{equation*}
R := \{v + v' \mid v' \in \mathrm{ker} f \}.
\end{equation*}
It remains to prove that $L = R$.
We do this in two steps:
first,  we prove $L \subset R$;
second, we prove $R \subset L$.
Please note: $v$ denotes an arbitrary input of $f$.
Step 1.
Let $x \in L$.
In this step, it remains to prove that $x \in R$.
Trivially, we have $f(x) = f(v)$.
Thus, $f(x) - f(v) = 0$.
Since $f$ is linear,
we have $f(x) + f(-v) = 0$,
and finally we have $f(x - v) = 0$.
So trivially, $x - v \in \mathrm{ker} f$.
Therefore, with respect to $R$, we may choose $v' = x - v$.
Hence, $R$ contains an element $v + v' = v + x - v = x$.
In other words, $x \in R$. This concludes Step 1.
Step 2.
Let $x \in R$.
In this step, it remains to prove that $x \in L$.
Trivially, we have $x = v + v'$ for some $v' \in \mathrm{ker} f$.
Thus, $f(x) = f(v + v')$.
Since $f$ is linear,
we have $f(x) = f(v) + f(v')$.
Since $v'$ is in the kernel, $f(v') = 0$.
Thus, $f(x) = f(v)$.
So trivially, $x \in L$.
This concludes Step 2.
QED
 A: That proof is fine. I'll make a few suggestions on style:


*

*Try to omit "trivially"; In step 1, for instance, "So trivially" could be "so by definition". 

*Try for brevity. 
a. "In this step, it remains to prove" -> "We'll show that"
b. "Since $f$ is linear ..." -> "By linearity, $f(x) + f(-v) = 0$, so $f(x-v) = 0$, so $x-v \in \ker f$. 


*Stick with the active voice when possible. ("It remains to prove" -> "We'll show").



Note: @Martin Brandenburg's very terse proof in the comments would be fine for someone who'd done a lot of math. If you're just learning to do proofs like this one, writing out the details is probably a good idea until you notice that the details always seem to be the same...at which point you can omit some details and condense others, etc., until your proofs are as terse as his. Doing so is a Good Thing, in the sense that you don't suffer from a forest-and-trees problem, where trivial proofs separate the main ideas by so much that it's hard to follow the thread of a larger argument. 
$~$
Here's a rewrite:
$~$
$\textit{Proof}$
Let
\begin{equation*}
L := f^{-1}\big( f(v) \big).\quad\quad\quad\quad
\end{equation*}
and
\begin{equation*}
R := \{v + v' \mid v' \in \mathrm{ker} f \}.
\end{equation*}
We'll show that $L \subset R$ and $R \subset L$. 
Step 1.
Let $x \in L$.
In this step, it remains to prove that $x \in R$.
By the definition of $L$, $f(x) = f(v)$, so $f(x) - f(v) = 0$. 
By linearity of $f$, $f(x) + f(-v) = 0$,
and $f(x - v) = 0$.
So $x - v \in \mathrm{ker} f$. Letting $v' = x-v$, we see that 
$x \in R$. 
Step 2.
Let $x \in R$. We'll show $x \in L$.
By the definition of $R$, there's a $v' \in \mathrm{ker} f$ such that 
$$
x = v + v'.
$$
Apply $f$ to both sides, and expanding by linearity, we have
\begin{align}
f(x) &= f(v + v')\\
&= f(v) + f(v') \\
&= f(v) + 0 \\
&= f(v),
\end{align}
so $f(x) = f(v)$ and $x \in L$. 
