Counterexamples Of The Infinite Dimensional Linear Space

Question

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?

Answer 1

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 $ACw = AC(\sum \beta_j w_j) = \sum \beta_j ACw_j = \sum \beta_j w_j = w.$

Answer 2

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 $\mathscr{A}\circ\mathscr{C}=\text{id}_W$.