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?


1 Answer 1


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.

  • $\begingroup$ I think what's interesting about the second example is not that it's infinite dimensional, but that the domain and codomain ($X$ and $U$) are the same. $\endgroup$ Commented Dec 16, 2016 at 15:41
  • $\begingroup$ @Omnomnomnom: Yes, I'll add a remark on that, thank you. $\endgroup$ Commented Dec 16, 2016 at 15:42

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