Injective implies surjective? I have a question on the injectivity of linear maps: In my book on functional analysis (Peter Lax) it is stated that if $M: X \to U$ is a bounded linear map ($X$ and $U$ are normed spaces) then one quotients out the nullspace of $M$ the map $M_0: (X/N_M) \to U$ is one-to-one. If $M$ is injective, then $N_M=\{0\}$. 
Then $X/N_M=X$. Is the the same as saying that (under the above mentioned conditions) $M$ is injective implies $M$ is surjective? 
 A: No. Take for example the injective map 
$$M\colon \mathbb R\to \mathbb R^2,\quad M(x)=(x, 0),$$ 
or, if you prefer an infinite-dimensional example, 
$$
M\colon \ell^2\to \ell^2,\quad M(x_1, x_2, x_3\ldots)=(0, x_1, x_2\ldots).$$ 
Both maps are injective but they are not surjective. (See edit below for more details). 
The flaw in your proof is here: if $M\colon X\to U$ is a linear map then $$M_0\colon X/N\to M(X)$$ is an injective map. Note the difference with what you wrote, which is that "$M_0\colon X/N\to U$ is injective". There is no guarantee that the image of $M_0$ is the whole of $U$, as the above examples show.
EDIT As Omnomnomnom points out, in the first example the domain and codomain of $M$ are different and finite-dimensional, while in the second they coincide and are infinite-dimensional. The reason for such a choice is that in the finite-dimensional case, there is a case in which linear maps are injective if and only if they are surjective. Namely, a linear map 
$$M\colon V_1\to V_2$$ 
where $V_1$ and $V_2$ are finite dimensional spaces of the same dimension is injective if and only if it is surjective. 
