Linear maps over finite and infinite dimensions. 
Let $ L, K: V \to V $ be linear maps that satisfy $L\circ K=1_V$. Show that 
  
  
*
  
*(a)  If $\ V $ is finite dimensional, then $ K\circ L=1_V$.
  
*(b)  If $\ V $ is infinite dimensional give an example where $K\circ L \neq 1_V $.
  

I am having trouble seeing why finite or infinite dimensions apply in these cases. 
 A: I will show a counterexample for the infinite dimensional case. Let $V$ be $\mathcal l_2 (\mathbb N)$, $K$ the right-shift operator, defined as $K(a_1, a_2, \ldots)=(0, a_1, a_2, \ldots)$ and $L$ the left-shift operator, defined as $L(a_1, a_2, \ldots)=(a_2, a_3, \ldots)$; they are both linear, and of course $LK=1_V$. It is also clear that $KL$ is not the identity. 
The statement is true in the finite dimensional case, because a surjective/injective endomorphism is also injective/surjective (for example, because it sends basis in basis), so it is invertible, and moreover the inverse is unique.
A: Hint: There's an analogous statement for sets: let $f,g:S\to S$ be functions with $f\circ g=\mathrm{id}_S$.
Then, $f$ must be surjective and $g$ injective. 
Moreover, for any $t=g(s)$ we have $g(f(t)) =g(f(g(s)))=g(s)=t$. 
But if $S$ is finite, simply by counting elements, both $f$ and $g$ will be bijective, so by the above, $g\circ f=\mathrm{id}_S$, 
while if $S$ is infinite, this conclusion does not hold. 
