Show that $L_{l}(X)$ is isomorphic to $L(X/Im(l),Ker(l))$

Question

Given two vector spaces $X$ and $Y$. The set of linear operators from $X$ to $Y$ is denoted by $$ L(X,Y) $$ if $Y=X$ then $$ L(X):=L(X,X) $$

Let $X$ be a vector space and let $l\in L(X)$, that is, $l:X\to X$ is a linear operator. $L_{l}(X)$ is a subspace of $L(X)$ given by $$ L_{l}(X):= \{ f\in L(X): f\circ l = 0, \ \ l\circ f =0 \} $$ I want to show that $L_{l}(X)$ is isomorphic to $L(X/Im(l),Ker(l))$. In other words, I want to show that there exists a bijective linear map $I:L_{l}(X)\to L(X/Im(l),Ker(l))$.

EDIT:

Consider the function $F$ given by \begin{align*} F: & L_{l}(X)\to L(X/Im(l);Ker(l))\\ & f\longmapsto \widehat{f} \end{align*} where $\widehat{f}:X/Im(l)\to Ker(l)$, is defined as follows $\widehat{f}(x+Im(l))=f(x)$.

Let $f_{1}$, $f_{2}\in L_{l}(X)$ such that $f_{1}=f_{2}$. we have $F(f_{1})(x+Im(l)) = f_{1}(x)$ and $F(f_{2})(x+Im(l)) = f_{2}(x)$, for every $x\in X$. We know that $f_{1}=f_{2}$ then $F(f_{1})=F(f_{2})$, for this reason $F$ is well defined.

Let $f_{1}$, $f_{2}\in L_{l}(X)$ such that $F(f_{1})=F(f_{2})$, this implies that for each $x\in X$ we have $F(f_{1})(x+Im(l)) = F(f_{2})(x+Im(l)) \implies f_{1}(x) = f_{2}(x)$, then $f_{1}=f_{2}$, that is, $F$ is injective.

Let $ g \in L(X/Im(l);Ker(l))$ I want to show that there exists $f\in L_{l}(X)$ such that $F(f) = g$. But I don't know how to proceed.

This is what I have done so far, but I don't know how to prove that $F$ is surjective. Any hints? Thanks in advance.

Answer 1

Your definition of $\hat{f}$ by $$\hat{f}(x+{\rm Im}\,l)=f(x)\in \ker l$$ is fine, but we have to show that the definition does not depend on $x.$ To this end let $x+{\rm Im}\,l=x'+{\rm Im}\,l.$ Then $x-x'\in {\rm Im}\,l.$ The condition $f\circ l=0$ implies that $f$ vanishes on ${\rm Im}\,l.$ Hence $f(x-x')=0,$ i.e. $f(x)=f(x').$

Now we need to show that the mapping $L_l(X)\ni f\mapsto \hat{f}\in L((X/{\rm Im}\,l),\ker l)$ is injective. Let $\hat{f}=0.$ Thus $\hat{f}(x+{\rm Im}\,l)=0$ for any $x\in X.$ This implies $f(x)=0$ for any $x,$ hence $f=0.$

It remains to show that $L_l(X)\ni f\mapsto \hat{f}\in L((X/{\rm Im}\,l),\ker l)$ is surjective. Let $g\in L((X/{\rm Im}\,l),\ker l).$ Define $f\in L(X)$ by $$f(x)=g(x+{\rm Im}\,l)\in \ker l.$$ As $f(x)\in \ker l$ for any $x\in X$ we get $l\circ f=0.$ Moreover $$(f\circ l)(x)= f(l(x))=g(l(x)+{\rm Im}\,l)=g(0+{\rm Im}\,l)=0$$ Hence $f\circ l=0.$ We have obtained that $f\in L_l$ and $\hat{f}=g.$