Let $V$ and $W$ be two finite dimensional linear spaces over the field $\mathbb{F}$ and $\mathscr{A} :V\rightarrow W$ be a linear map between $V$ and $W$. Then we have

  1. $\mathscr{A}$ is an injective linear map if and only if there exists a linear map $\mathscr{B}:W\rightarrow V$ such that $\mathscr{BA}=\mathit{Id}_{\mathbf{V}}$. ($\mathit{Id}_{\mathbf{V}}$ is the identity map on $V$)

  2. $\mathscr{A}$ is a surjective linear map if and only if there exists a linear map $\mathscr{C}:W\rightarrow V$ such that $\mathscr{AC}=\mathit{Id}_{\mathbf{W}}$.

I think that 1 and 2 above don't hold if $V$ and $W$ are infinite dimensional linear spaces over $\mathbb{F}$.

I need some counterexamples to verify my idea. How can I find them?

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    $\begingroup$ Just define them on a basis as you would do for finite dimensional spaces. $\endgroup$ – Ennar Oct 1 '18 at 12:42
  • $\begingroup$ @Ennar:You mean both $(3)$ and $(4)$ are right conclusions? $\endgroup$ – user553010 Oct 1 '18 at 12:52
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    $\begingroup$ This could be helpful to apply Ennar's suggestion : math.stackexchange.com/questions/1351848/… $\endgroup$ – Arnaud D. Oct 1 '18 at 12:57
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    $\begingroup$ Note however, that the argument suggested by @Ennar only applies if you are willing to use the axiom of choice, which is needed to get the standard results on bases needed to do this in infinite dimensions. This also implies that it is impossible to construct explicit counter examples. $\endgroup$ – Andreas Cap Oct 1 '18 at 13:00
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    $\begingroup$ @Andreas is right since "every vector space has a basis" is equivalent to the axiom of choice. $\endgroup$ – Ennar Oct 1 '18 at 13:02

Exercise (1):

$\Leftarrow$: Assume $A$ was not injective, i.e. we had $v, v'$ with $Av = Av'$. Then $BAv = v = v' = BAv'$, which is a contradiction.

$\Rightarrow$: Define the linear map $B: Im(A) \to V$ by

$$B(w) \in A^{-1}(w)$$

for every $w \in Im(A)$. This definition is unique and well-defined since $A$ is injective. $B$ is as the (effective) inverse of a linear map also linear.

Extend $B$ from the subspace $Im(A)$ to $W$ linearly. Then obviously $BAv = B(Av) = v$ by definition.

You can indeed extend $B$ linearly by noting that within vector spaces every subspace has a complement (e.g. see here): There is a subspace $M$, so that $W = Im(A) \oplus M$. Now define $B': W \to V$ by $B'(w) = B'(x + y) := B(x) + 0$ with $x \in Im(A)$ and $y \in M$.

Alternative: Let $\{v_i | i \in I\}$ be a basis of $V$. Then $N := \{Av_i | i \in I\}$ is a set of linearly independent vectors as well:

$$0 = \sum_k \alpha_k Av_k \Rightarrow A^{-1}0 = A^{-1} \sum_k \alpha_k Av_k \Rightarrow 0 = \sum_k \alpha_k v_k \Rightarrow \forall k. \alpha_k = 0$$

Extend $N$ to a basis of $W$, namely $N \cup \{w_j | j \in J\}$ and define $B: W \to V$ using Ennar's suggestion by its image on the basis elements: $B(Av_i) = v_i$ and $B(w_j) = 0$ (arbitrary).

Then $BAv = BA(\sum_k \alpha_k v_k) = \sum_k \alpha_k B(Av_k) = \sum_k \alpha_k v_k = v$ as required.

Exercise (2)

$\Leftarrow$: $AC = Id_W$ directly implies that $A$ is surjective.

$\Rightarrow$: Again, using Ennar's suggestion , Define the linear map $C: W \to V$ by $$C(w_i) \in A^{-1}(w_i).$$ Then $ACw = AC(\sum \beta_j w_j) = \sum \beta_j ACw_j = \sum \beta_j w_j = w.$

  • $\begingroup$ Since $Im(A)\subset W$,$W$ instead of $V$ in $V = Im(A) \oplus M$ should be right.Obviously,It was just a slip of the open. Your answer expressed what I meant to say.+1 $\endgroup$ – user553010 Oct 2 '18 at 13:48
  • $\begingroup$ @user553010 Thanks! It's now corrected. $\endgroup$ – ComFreek Oct 2 '18 at 15:41

In the language of category theory, for a division ring $\mathbb{F}$, every object in the abelian category $\mathcal{C}:=\mathbf{Vec}(\mathbb{F})$ of $\mathbb{F}$-vector spaces and $\mathbb{F}$-linear maps is both injective and projective. That is, any exact sequence $$0\to A \to B\to C\to 0$$ of objects and morphisms in $\mathcal{C}$ splits. In other words, the morphism $A\to B$ has a left inverse $B\to A$, the morphism $B\to C$ has a right inverse $C\to B$, and $B\cong A\oplus C$. The proof relies on the Axiom of Choice in creating a basis, or extending a linearly independent subset to a basis of a vector space.

How does this solve your problem? Well, if $\mathscr{A}:V\to W$ is an injective map, then there exists a short exact sequence $$0\to V\overset{\mathscr{A}}{\longrightarrow} W \to W/\text{im}(\mathscr{A})\to 0\,,$$ which splits, whence there exists a left inverse $W\overset{\mathscr{B}}{\longrightarrow} V$ of $\mathscr{A}$, making $\mathscr{B}\circ\mathscr{A}=\text{id}_V$. If $\mathscr{A}:V\to W$ is a surjective map, then there exists a short exact sequence $$0\to \ker(\mathscr{A}) \to V\overset{\mathscr{A}}{\longrightarrow} W\to 0\,,$$ which splits, whence there exists a right inverse $W\overset{\mathscr{C}}{\longrightarrow} V$ of $\mathscr{A}$, making $\mathscr{A}\circ\mathscr{C}=\text{id}_W$.


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