Proof: $\exists$ subspace $U$ of $ker(f)$ with $U \bigoplus T_1 = T_2 $ I need help with this proof:
Let $V, W$ be K-vectorspaces.
Let $T_1, T_2$ be subspaces of V with $T_1 \subseteq T_2$.
Let $f \in hom_K(V,W)$.
Show the following: If $ f(T_1) = f(T_2)$ then exists a subspace $U$ of $ker(f)$ so that $ U \bigoplus T_1 = T_2$
I'm sorry, but I really have no clue how to begin at least one direction of this proof, so I'm grateful for any kind of help!
 A: Put $U_1:=T_1\cap {\rm ker}(f)$, $\>U_2:=T_2\cap {\rm ker}(f)$. It follows that $U_1\subset  U_2\subset{\rm ker}(f)$. 
Let $U$ be a complement of $U_1$ in $U_2$, so that we have
$$U_2=U_1\oplus U\ .\tag{1}$$
I claim that
$$T_2=T_1\oplus U\ .\tag{2}$$
Proof. As $T_1\subset T_2$ and $U\subset U_2\subset T_2$ we trivially have $T_1+U\subset T_2$.
On the other hand, consider an arbitrary $x\in T_2$. By assumption on $f$ there is an $x'\in T_1$ with $f(x')=f(x)$. It follows that $y:=x-x'\in{\rm ker}(f)$, and as $y\in T_2$ as well we have $y\in U_2$. Using $(1)$ we now can find a $y'\in T_1$ and a $u\in U$ with $y=y'+u$. In all  we obtain
$$x=x'+y=(x'+y')+u\in T_1+U\ ,$$
which proves the converse inclusion $T_2\subset T_1+U$.
In order to prove that the sum $(2)$ is direct consider an $x\in T_1\cap U$. As $U\subset U_2\subset {\rm ker}(f)$ it follows that $x\in T_1\cap{\rm ker}(f)=U_1$. Since the sum $(1)$ is direct we conclude that $x=0$.
A: Not a full answer: We have that $f(T_2)=f(T_1)$ hence in particular we have the first inclusion $f(T_2)\subset f(T_1)$. For each $t_2\in T_2$ we have that $f(t_2)\in f(T_2)\implies f(t_2)\in f(T_1)$, then there  exists $t_1\in T_1$ such that 
 $$f(t_2)=f(t_1)\implies f(t_2-t_1)=0\implies t_2-t_1\in \ker f$$ 
Hence there exists $u\in \ker(f)$ such that 
$$t_2-t_1=u\implies t_2=u+t_1$$
Hence 
$$T_2=\ker f+T_1$$
Now using the other inclusion $f(T_1)\subset f(T_2)$ we show in a similar way that
$$T_1=\ker f+T_2$$.
