Say $f:V\rightarrow U$ is a linear map with $\ker(f)=W$ and $f(v)=u$. Show that the set $v+W=\{v+w:w\in W\}$ is the preimage of $u$.
My proof:
$v \in W:f(v)=u$, $w\in W\implies f(w)=0$
$f(v+w)=u$ by linearity, $f^{-1}(u)=f^{-1}(f(u+w))=v+w$, $\forall w \in W$, by property of inverse.
This imples $f^{-1}(u)=v+W$
Proof in book: Suppose $v'\in f^{-1}(u)$ then $f(v')=u$ and so $f(v'-v)=f(v')-f(v)=u-u=0$, that is $v'-v\in W$ thus $v'=v+(v'-v)\in v+W$, and hence $f^{-1}\subseteq v+W$.
Suppose $v'\in v+W$ then $v'=v+w$ where $w\in W$ this implies $f(v')=f(v+w)=f(v)+f(w)=f(v)=u$ thus $f^{-1}(u)$ and so $v+W\subseteq f^{-1}(u)$ both inclusions imply $f^{-1}=v+w$.
Does my proof work? Or am I making some assumption that I am not able to see.
You proof is wrong. When you write that “$f^{-1}(u)=f^{-1}(f(u+w))=v+w$, $\forall w \in W$, by property of inverse”, you act as if $f$ has an inverse. If $W\neq\{0\}$ it hasn't. The notation $f^{-1}(v)$ denotes the set of the elements $u\in V$ such that $f(u)=v$. In other words, $f^{-1}$ is used here for the inverse image of the set $\{v\}$.