Is $\ker(T)=U\cap W$ right? 
Let $U,W$ be the subspaces of a vector space $V$. The map of $U\times W: \to V$ is given by: $(u,w)\to u-w$. Show that the kernel is isomorphic to $U\cap W$.

$T:U\times W\to V$
$T(u,w)=0\:\:\text{iff}\:\:u=w,u\in U,w \in W$
$u=w\:\:\text{iff}\:\:u,w\in U\cap W$
So $\ker(T)={u,w\in U\times W:f(u,w)=0}=U\cap W$
So $\ker(T)=U\cap W$
Therefore, $\dim(ker(T))=\dim(U\cap W)$ which means $T$ is surjective and since $\ker(T)=U\cap W$ it is obviously injective, therefore it is an isomorphism.
Questions:
1) Is $\ker(T)=U\cap W$ right?
2) If $\ker(T)\neq U\cap W$, How do I build a map $T´:\ker(T)\to U\cap W$ and prove $T´$ to be an isomorphism?
Thanks in advance!
 A: $1.$ No, because $\ker T\subset U\times W\subset V\times V$ whereas $U\cap W\subset V$. 
However it is isomorphic to $U\cap W$, since $\ker T=\{(u,w)\mid u\in U,\;w\in W, \;u=w\}=\{u,u)\mid u\in U\cap W\}$, whence the inverse isomorphisms:
$$\begin {align}
\ker T&\longrightarrow U\cap W&&&U\cap W&\longrightarrow \ker T\\
(u,u)&\longmapsto u&&&u&\longmapsto(u,u) \\
\end{align}$$
A: It is not true that $u = w$ iff $u,w\in U \cap W$.  You've made some strange/mistaken statements after that.  Certainly, we cannot say that $\ker(T) = U \cap W$, since $U \cap W$ is not a subspace of $U \times V$ (the domain of $T$).
Here's what we can say:
$$
\ker(T) = \{(u,w) : u \in U, w \in W, u - w = 0\} \\
= \{(u,w):u \in U, w \in W, u = w\}\\
= \{(u,u): u \in (U \cap W)\}
$$
That is, the kernel of $T$ is the set of all "repeating" pairs of the form $(u,u)$ where $u$ is taken from $U \cap W$.  We want to show that this subspace of $U \times W$ is isomorphic to $U \cap W$.
Now, once you've understood what the space $\ker(T)$ looks like, there is an "obvious" isomorphism $\phi:\ker(T) \to U \cap W$.  Can you figure it out?
