For a given transformation $T: V \rightarrow W$ why does $im(T) \cong V/ker(T)$ I was looking at this example on Wikipedia, but unfortunately I don't quite see the connection. I am mostly trying to understand it through this picture:

Does anyone have a better explanation? I think the main part that is throwing me off is the exclusion of ker(T).
 A: For any $w\in W$, any two elements of $T^{-1}(w)$ 'differ by' an element of the kernel.  That is, let $w=T(v_1)=T(v_2)$.  Then by linearity $T(v_2-v_1)=0$.  In other words, $v_2-v_1\in\rm{ker}T$.
Conversely, if you add any element of the kernel to a particular solution of $Tx=w$, you get another solution.
This may help with some intuition.
Incidentally, this one has a special name:  the first isomorphism theorem.
A: That's probably the worst diagram on the planet for trying to visualize the first isomorphism theorem for vector spaces.
Instead, think about ordinary 3-dimensional Euclidean space $\mathbb{R}^3$. The subspaces are: the origin (zero), lines through the origin, planes through the origin, and the whole space.
A linear map $L:\mathbb{R}^3\to\mathbb{R}^3$ always "kills off" a subspace (i.e. maps a subspace to the origin), which is called the kernel. Whatever it doesn't kill off is in the image subspace. For example, projection onto the $xy$-plane kills off the $z$-axis.
Taking a quotient by a subspace kills off that subspace. For example if $U$ is the $z$-axis, then $\mathbb{R}^3/U$ is (isomorphic to) the $xy$-plane.
This should hopefully help you understand the theorem.
