how can I prove that $\operatorname{im}(K \circ L) \subseteq \ker(K)$? I am struggling with an algebra problem here is what we got :
Assume we have :
$$L : E_0 \to E_1 \quad\text{and}\quad K : E_1 \to E_2 $$
We have to show that :
$$\dim(\ker(K \circ L)) \leq \dim(\ker(L)) + \dim(\ker(K)).$$
I define $H$ as :
$$H : \ker(K \circ L) \to E_2, \qquad H(v) = L(v) \quad \text{for} \quad v ∈ \ker(K \circ L)$$
Now I have to prove that :
$$ \ker(L) = \ker(H) $$
But how can I show that :
$$\Rightarrow \qquad \ker(L) \subseteq \ker(K \circ L) \quad \text{and} \quad \ker(K \circ L) \subseteq \ker(L) \quad ?$$
I got an idea for this one : $L(v) = 0$ implies $K(L(v)) = 0$ ... But I stuck with this one.
$$ \Rightarrow \qquad \operatorname{im}(H) \subseteq \ker(K) \quad ? $$
How can I prove? I only know and prove that $\operatorname{im}(K\circ L) \subseteq \operatorname{im}(K)$.
Thanks.
 A: Generally you cannot show that $\ker (\mathcal {KL}) = \ker \mathcal L$: if $\mathcal Kv =0$ for all $v \in E_1$, where $E_1 \neq 0$, and $\mathcal L$ surjective, then $\ker(\mathcal {KL}) = E_0$ but $\ker \mathcal L\subsetneq E_0$ [otherwise $\mathcal L$ would not be surjective: some $v \in E_1 \setminus \{0\}$ cannot be mapped by $\mathcal L$]. 
Assume $\dim(E_0) = n <+\infty$. By the Rank-Nullity Theorem, $ \dim(\ker \mathcal L)+ \dim(\mathrm {im}\, \mathcal L) = n$,  also $
\dim(\ker (\mathcal {KL})) + \dim(\mathrm {im} (\mathcal {KL})) =n
$.
Then 
\begin{align*}
\dim(\ker (\mathcal {KL})) &= n - \dim (\mathrm {im} (\mathcal {KL}))\\
&= \dim (\ker \mathcal L) + \dim(\mathrm {im} (\mathcal L )) \\
&\quad - \dim (\mathrm {im}(\mathcal {KL})). 
\end{align*}
Note that the mapping $\mathcal K\vert _{\mathrm {im}\,\mathcal L} \colon \mathrm {im}\, \mathcal L \to E_2$ is a linear mapping, and we could know that $\mathrm {im}(\mathcal K\vert _{\mathrm {im}\, \mathcal L}) = \mathrm {im} (\mathcal {KL})$: $z \in \mathrm {im} (\mathcal K\vert _{\mathrm {im} \,\mathcal L}) \iff \exists y \in \mathrm {im}\, \mathcal L, \mathcal Ky = z \iff \exists x \in E_0, y = \mathcal Lx, z = \mathcal Ky \iff \exists x \in E_0, z = \mathcal {KL}x$. So again by the Rank-Nullity Theorem, 
$$
\dim (\mathrm {im}\, \mathcal L) = \dim(\ker (\mathcal K\vert _{\mathrm {im}\,\mathcal L})) + \dim (\mathrm {im}(\mathcal {KL})). 
$$
Thus
$$
\dim (\mathrm {im}\, \mathcal L) -  \dim (\mathrm {im}(\mathcal {KL})) =\dim(\ker (\mathcal K\vert _{\mathrm {im}\,\mathcal L})) \leqslant \dim (\ker \mathcal K), 
$$
the $\leqslant$ holds because $\mathrm {im} \,\mathcal L \subset E_1$, and possibly there is some $y \in (E_1\setminus \mathrm {im}\, \mathcal L)$ that $\mathcal Ky = 0$. 
Hence the inequality we want to prove. 
