Proof ${\rm Ker}(L○T) = {\rm Ker}(T) \iff {\rm Im}(T) \cap {\rm Ker}(L) = \{0\}$ Let $V,W$ be two vector spaces with scalars $K$,
and $T : V  \rightarrow W$ and $L : W  \rightarrow W$
two linear transformations.
Prove that $\def\Ker{\operatorname{Ker}}\Ker(L \circ T) = \Ker(T)$ if and only if $\def\Im{\operatorname{Im}}\Im(T) \cap \Ker(L) = {0}$

a) $\Ker(L \circ T) = \Ker(T) =  \{\,\vec{v} \in V : T(\vec{v})=L(T(\vec{v})) = 0\,\}$
Then $\Ker(L) = 0$ (It feels right, but don't know how to explain this step)
$L(\Im(T))=0$, then $\Im(T) \subset \Ker(L) = \{0\}$
$\Im(T) \cap \Ker(L) = \{0\}$

b) $\Im(T) \cap\Ker(L) = \{0\}$
$L(\Im(T))=\{0\}$, then $\Im(T) \subset \Ker(L)$
but, $\Im(T) \cap \Ker(L) = {0}$, then 
$\Ker(L) = 0$
$\Ker(L \circ T) = \Ker(T)$
This is the best I could come up with, and I myself don't like it.
In my head, there's this idea, that  $\Ker(L \circ T)$ is equal to $\Ker(L)+\Ker(T)$ (that's why at some point I do $\Ker(L) = \{0\}$) , but at the same time I know the vectors of $\Ker(L)$ and $\Ker(T)$ don't really have to be living in the same space, then, such addition doesn't make sense.
 A: Your proof for $a$ and you proof for $b$ are wrong. An explanation below (I explain why you are wrong, also I give a different correct proof).
Let's begin from the start you have:
$\ker(L \circ T) = \ker(T) =  \{\vec{v} \in V : T(\vec{v})=L(T(\vec{v})) = 0\}$
The claim that $\ker (L)=0$ is false in general. For instance if $T=L$ are the zero maps.
Heres is how to do it: Take $w\in \ker(L)\cap Im(T)$. Then there exists $v\in V$ such that $w=T(v)$.
$L(w)=0$ implies that $L(T(v))=0$ which implies that $v\in \ker L\circ T = \ker T$ hence, $T(v)=0$, we conclude that $w=0$. Therefore $\ker(L)\cap Im(T)=\{0\}$.
Now for the other direction:
We have, $Im(T) \cap \ker (L)=0$. This does not imply that $L(Im(T))=0$. For instance if $V=\mathbb{R}$, $W=\mathbb{R}^2$, $T(x)=(x,0)$ and $L(x,y)=(x-y,x-y)$. Then $Im(T) = \{(x,0):x\in\mathbb{R}\}$ while $\ker(L) = \{(z,z):z\in\mathbb{R}\}$ and so the intersection is zero, however $L(Im(T)) = Im(T)\not=0$.
Another way to go is as follows: First observe that $\ker(T)\subseteq \ker (L\circ T)$ (this is trivial $Tv=0\Rightarrow L(T(v))=L(0)=0$). It is therefore enough to prove inclusion in the other direction.
For this, let $v\in \ker L\circ T$, so we have $L(T(v))=0$. This implies that $T(v)\in \ker (L)$, and clearly $T(v)\in Im(T)$. As $Im(T)\cap\ker L = 0$ we have that $T(v)=0$ therefore $v\in\ker T$. This proves inclusion in the other direction.
A: In both directions you make essential errors. For that see the answer of Yanko.

It is evident that $\mathsf{ker}T\subseteq\mathsf{ker}(L\circ T)$.
This on base of $T(x)=0\implies L(T(x))=0$.
So the following statements are equivalent:


*

*$\mathsf{ker}(L\circ T)\neq\mathsf{ker}T$.

*Some $x$ exists with $x\in\mathsf{ker}(L\circ T)$ and $x\notin \mathsf{ker}T$.

*Some $x$ exists with $L(T(x))=0$ and $T(x)\neq0$.

*Some $y\in\mathsf{im}T$ exists with $L(y)=0$ and $y\neq0$.

*Some $y\in\mathsf{im}T$ exists with $y\in\mathsf{ker}L$ and $y\neq0$.

*$\mathsf{im}T\cap\mathsf{ker}L\neq\{0\}$.


This allows us to conclude: $$\mathsf{ker}(L\circ T)\neq\mathsf{ker}T\iff\mathsf{im}T\cap\mathsf{ker}L\neq\{0\}$$or equivalently:$$\mathsf{ker}(L\circ T)=\mathsf{ker}T\iff\mathsf{im}T\cap\mathsf{ker}L=\{0\}$$q.e.d.
A: I can't really follow any of your reasoning. But here is how you can argue.
The left hand side only mentions $\def\Ker{\operatorname{Ker}}T$ and $L\circ T$, so it does not "see" anything in $W$ outside the image $\Im(T)$. It is therefore natural to consider the restriction $\def\Im{\operatorname{Im}}L|_{\Im(T)}$ of $L$ to $\Im(T)$, which I will abbreviate to $L':\Im(T)\to W$. Since all vectors in $\Im(T)$ lie by definition in the domain of $L'$ we can apply $L'$ after $T$, to obtain the same map $V\to W$ as $L\circ T$, namely $v\mapsto L(T(v))$. To view this as a proper composition with $L'$, we need to modify the codomain of $T$, so let $T':V\to\Im(T)$ be the surjective map given by $v\mapsto T(v)$ (so it is just $T$ with a restricted codomain), then one has equality of maps $L'\circ T'=L\circ T$.
So in particular $\Ker(L\circ T)=\Ker(L'\circ T')$.
For the right hand side, we can remark that $\Im(T)\cap\Ker(L)=\{\, w\in\Im(T)\mid L(w)=0\,\} = \Ker(L')$.
So now we are asked to prove for a surjective linear map $T':V\to U$ and a linear map $L':U\to W$ that $\Ker(L'\circ T')=\Ker(T')\iff\Ker(L')=\{0\}$. On both sides the inclusions $\supseteq$ are always true, so we may prove just $\Ker(L'\circ T')\subseteq\Ker(T')\iff\Ker(L')\subseteq\{0\}$. Assume the left hand inclusion and let $w\in\Ker(L')$; by surjectivity we can write $w=T'(v)$ for some $v\in V$; then $0=L'(w)=L'(T'(v))$ so that $v\in\Ker(L'\circ T')\subseteq\Ker(T')$ and therefore $w=T'(v)=0$ as desired. Conversely assume $\Ker(L')\subseteq\{0\}$ and let $v\in\Ker(L'\circ T')$; then $L'(T'(v))=0$ so $T'(v)\in\Ker(L')\subseteq\{0\}$ and $v\in\Ker(T')$ as desired.
