Doubts about fundamental theorem of Homomorphism 
(Fundamental Theorem of Homomorphism) : Let $T$ be a linear transformation of a vector space $V$ onto a vector space $U$ over the field $\mathbb{F}$. Then 
  $$V/\ker T \cong U$$

I doubt that a mapping that was onto in the first place becomes bijective when considered $V/\ker T \rightarrow U$. I have read 
this wikipedia article about kernels and it states that "the kernel of a homomorphism measures the degree to which the homomorphism fails to be injective". That is, by changing $V$ to $V/\ker T$ we are somehow making the mapping one-to-one. By taking the quotient space, it somehow takes care of the elements which are mapped to the same element in $U$.
How this is possible - any examples or insight please?
 A: Let’s simplify the situation for a moment. Forget that $V$ and $U$ are vector spaces and that $T$ is a linear transformation; just think of $V$ and $U$ as sets and of $T$ as a map from $V$ onto $U$. For each $u\in U$ let $V_u=\{v\in V:T(v)=u\}$, the set of points in $V$ that are mapped to $u$ by $T$. Let $\mathscr{P}=\{V_u:u\in U\}$. If $u_0,u_1\in U$ and $u_0\ne u_1$, then $V_{u_0}\cap V_{u_1}=\varnothing$: $T$ cannot send any $v\in V$ both to $u_0$ and to $u_1$. Thus, the map $\varphi:U\to\mathscr{P}:u\mapsto V_u$ is a bijection, and so, of course, is its inverse sending $V_u\in\mathscr{P}$ to $u$.
Now put the linear algebra back into the picture. First, $\ker T=\{v\in V:T(v)=0_U\}$, so in the notation of my first paragraph, $\ker T=V_{0_U}$: it’s one of the members of $\mathscr{P}$. Fix $v_0\in V$ and let $u_0=T(v_0)$; what vectors in $V$ belong to $V_{u_0}$? Suppose that $v\in V_{u_0}$; then $T(v)=u_0=T(v_0)$. Since $T$ is linear, $T(v-v_0)=T(v)-T(v_0)=0_U$, so $v-v_0\in\ker T$, and $v\in v_0+\ker T$, where $v_0+\ker T=\{v_0+v:v\in\ker T\}$. Conversely, you can easily check that if $v\in v_0+\ker T$, then $T(v)=u_0$, and therefore $v\in V_{u_0}$. Thus, $V_{u_0}=v_0+\ker T$. In other words, the members of $\mathscr{P}$ are precisely the sets of the form $v_0+\ker T$ for $v_0\in V$.
By definition the members of $V/\ker T$ are the sets $v_0+\ker T$ for $v_0\in V$, and we’ve just seen that these are the members of $\mathscr{P}$, so in fact $V/\ker T=\mathscr{P}$. Thus, we can just as well think of the map $\varphi$ defined above as a bijection from $U$ onto $V/\ker T$. Its inverse, which I’ll call $h$, is a bijection from $V/\ker T$ onto $U$. What does $h$ look like? Let $v_0+\ker T\in V/\ker T$, and let $u_0=T(v_0)$. We’ve just seen that $v_0+\ker T=V_{u_0}$, and we know from the first paragraph that $h(V_{u_0})=u_0$. In other words, $h(v_0+\ker T)=u_0=T(v_0)$. 
We’ve now shown that the map $h:V/\ker T\to U:v+\ker T\mapsto T(v)$ is a bijection; in terms of the sets involved, it’s just the inverse of the bijection $\varphi$ of the first paragraph. To complete the proof that $V/\ker T$ and $U$ are isomorphic, we just check that $h$ is linear, which is a straightforward computation.
A: An example:
Consider the projection $p\colon \mathbb{R}^2\rightarrow\mathbb{R}$, defined by $p(x,y) = x$. This map is linear, surjective, but not injective, so it satisfies your assumptions. Here $V=\mathbb{R}^2$ and $U=\mathbb{R}$.
For every $x_0\in\mathbb{R}$, the whole line $\{(x_0,y)~|~y\in\mathbb{R}\}$ is mapped to $x_0$. So for every point in the image we have a whole line in the domain. If we want to make $V$ "smaller", such that $p$ becomes injective, we have to collapse each of these lines to a single point, namely the $x$-coordinate.
That is, we should consider only $\mathbb{R}$ as the domain for a function $\tilde{p}$ that is similar to $p$, to make $\tilde{p}$ injective. It will be surjective, because we have not removed too many points.
What does this have to do with the kernel and the algebraic construction? Well, what is the kernel of $p$? It is the line that is mapped to $0$:
$$\mbox{ker}(p) = \{(x,y)\in\mathbb{R}^2~|~p(x,y)=0\} = \{(0,y)~|~y\in\mathbb{R}\}$$
Consider $V/\mbox{ker}(p)$. This is the vector space $V=\mathbb{R}^2$, but with two points identified, if their difference is in $\mbox{ker}(p)$. That is two points $(x_1,y_1)$ and $(x_2,y_2)$ will be considered "the same", if
$$(x_1,y_1)-(x_2,y_2)\in\mbox{ker}(p) \Leftrightarrow x_1-x_2 = 0 \Leftrightarrow x_1=x_2$$
We see, that this construction gives exactly what we wanted: All the points on the line $\{(x_0,y)~|~y\in\mathbb{R}\}$ are identified, so the line collapses into one point.
A: The quotient $V/\ker T$ is defined putting some relation between the elements of $V$, that is, two classes $\bar x,\bar y$ are the same if they can be represented by elements such that $x-y\in \ker T$. So, if you have two elements $u,v\in V$ such that $T(u)=T(v)$, passing to the quotient these elements are the same and so your induced map (that is, $T':V/\ker T\to U$) would be injective.
A: Basically, the idea is that for homomorphisms, non-injectivity is "nicely behaved": Given two vectors spaces $V, W$ and a homomorphism $f: V \to W$, the set of vectors $f^{-1}(\{w\})$ that map to $w$ (i.e., $f^{-1}(\{w\}) := \{ v | f(v) = w \}$) forms an affine space, i.e., there is some vector space $V' \subseteq V$ such that $f^{-1}(\{w\}) = v + V'$ for any $v \in V$ such that $f(v) = w$. Furthermore, one finds that $V'$ is always given by $\ker f$: If $v, v' \in V$ and $f(v) = f(v')$, then $v - v' \in \ker f$. This follows from linearity of $f$.
Now, what is the vector space $V/\ker f$? It is the vector space of all classes of vectors $[v]$ where $[v] = [v']$ iff $v - v' \in \ker f$. Now, let $\bar f([v]) := f(v)$ (ignoring well-definedness for now). I claim that $\bar f$ is injective. To see this, assume $\bar f([v]) = \bar f([v'])$. Then $f(v) = f(v')$, so $v-v' \in \ker f$ (see above), and thus, $[v] = [v']$. In other words, $\bar f$ is injective. If $f$ is surjective, $\bar f$ is surjective too. Thus, we have found an isomorphism $\bar f: V/\ker f \to W$.
