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Let $V,~W$ be two vector space of a field $F$. Let $T:V\to W$ be a linear transformation. Then does $T$ has a left inverse linear transformation? I'd tried very hard to think of it, but I still can't see the way. And if so, can we prove it without Axiom of Choice?

PS: The problem is that, the left inverse $S$, must be a function such that $S\big\vert_{\text{range}(T)}(y)$ be defined by the value $x\in V$, such that $T(x)=y$. However, how do we define the function value of $S$ outside $\text{range}(T)$, such that $S$ is linear?

José Carlos Santos
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Eric
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  • This is a general propery of functions: $f$ is 1-to-1 iff $f$ has a left inverse. To show this, define the left inverse explicitly. In this case, the linearity is quite natural. – Ranc Sep 26 '17 at 17:00
  • @Ranc See my edit. I have a big trouble to show the linearity :(... need help! Thanks. – Eric Sep 26 '17 at 17:01
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    If $T$ is the inclusion of $F=V=\mathbb Q$ in $W=\mathbb R$, and a left inverse exists, then I believe its kernel gives you a Vitali set. (Haven't gone through the details.) So you need something beyond ZF. – Chris Culter Sep 26 '17 at 17:03
  • Ok, after seeing your edit: $\text{range}(T)$ is a linear space, and we need to send everything that is outside of $\text{range}(T)$ to elements in a way that doesn't spoil the linearity. The solution I'm thinking of is too simple so there aren't any non-revealing hints... so I'll just say it: Does $S(W\setminus \text{range}(T) ) = { 0 }$ work? @ChrisCulter, what do you think? – Ranc Sep 26 '17 at 17:06
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    @Ranc No, we need for example $S(\pi+1)-S(\pi)=1$. – Chris Culter Sep 26 '17 at 17:07
  • @ChrisCulter pshhhh.. thanks. This question just became very interesting to me. – Ranc Sep 26 '17 at 17:08

2 Answers2

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This statement is equivalent to the axiom of choice. Indeed, assuming the axiom of choice, you can take any basis $B$ of $V$, and then extend the linearly independent set $T(B)\subset W$ to a basis $C$ of $W$. Then define $S_0:C\to V$ by $S_0(c)=b$ if $c=T(b)$ for some $b\in B$ and $S_0(c)=0$ if $c\in C\setminus T(B)$. Since $C$ is a basis for $W$, this $S_0$ extends uniquely to a linear map $S:W\to V$, and this extension is a left inverse to $T$ since $S(T(b))=b$ for any $b\in B$ and $B$ spans $V$.

Conversely, suppose any linear injection has a linear left inverse. Then for any vector space $W$ and any subspace $V\subseteq W$, the inclusion map $T:V\to W$ has a left inverse $S:W\to V$. The kernel of $S$ is then a linear complement of $V$ in $W$: that is, a subspace $U\subseteq W$ such that $U\cap V=0$ and $U+V=W$. The existence of such a subspace in general is equivalent to the axiom of choice; see Existence of vector space complement and axiom of choice.

Eric Wofsey
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The space $T(V)$ has a complement $W^\star$ in $W$ (assuming the axiom of choice). So, if $T$ is one-to-one, you define $S:T(V)\longrightarrow V$ as its inverse, which is linear, and then you extend it to $W$ imposing, for instance, that $S|_{W^\star}\equiv0$. Then $S$ is a left-inverse of $T$ which is linear.

José Carlos Santos
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    Hmm.. seems the axiom of choice is inevitable right? Can we prove that we cannot throw away the axiom of choice to show this result? – Eric Sep 26 '17 at 17:19
  • @Eric I don't know, but I don't think so. – José Carlos Santos Sep 26 '17 at 17:29
  • Apparently it does require full AC: https://math.stackexchange.com/questions/1713561/existence-of-vector-space-complement-and-axiom-of-choice/ – Chris Culter Sep 26 '17 at 17:40
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    And the statement is indeed equivalent to the existence of complementary subspaces: if $W \subseteq V$ is a subspace and $\pi$ is a left inverse of the inclusion map $W \hookrightarrow V$, then the kernel of $\pi$ is a complementary subspace of $W$. – Daniel Schepler Sep 26 '17 at 17:44
  • By the way, as you said, $S\big\vert_{W\setminus\text{range}(T)}:=0$, why is $S$ linear here? I think it is trival at the first glance, but found it weird after thinking deeply. – Eric Sep 26 '17 at 17:44
  • @Eric I did not say that. What I wrote is that $W^\star$ is a complement of $T(V)$ in $W$ (that is, a vector subspace such that $T(V)\oplus W^\star=W$ and that $S|_{W^\star}\equiv0$. – José Carlos Santos Sep 26 '17 at 17:47