Kernels, linear maps and their compositions. I am reading through a paper where the author states what seems like a trivial fact. I would like to get my head around it. 
Let $V_1,V_2,W$ be vector spaces. 
Let $\alpha:V_1\rightarrow W$ and $\beta:V_1\rightarrow V_2$ be linear maps.
Is the following claim true? 
If $\ker(\alpha)=\ker(\beta)$, then there exists a linear map $\lambda:V_2\rightarrow W$ such that $\alpha=\lambda\circ \beta$.
How would I prove this? Are we able to construct this map explicitly?
The $\ker(\alpha)=\ker(\beta)$ is necessary since if the kernels were different and one took $v\in \ker(\beta)$ that didn't lie in $\ker(\alpha)$ then $(\lambda\circ\beta)(v)=0\neq \alpha(v)$. But is it sufficient? I feel there should be some easy way to construct the map?
Or am I missing something here and it's not actually true? I don't know if it is relevant - but the map $\beta$ is onto - if that helps. But I don't see how we could use that.
 A: We really need $\beta$ to be onto. Take $v_2\in V_2$ and define $\lambda(v_2) = \alpha(v_1)$, where $v_1 \in V_1$ is such that $v_2 = \beta(v_1)$. We can pick $v_1$ because $\beta$ is onto.
The map $\lambda$ is well defined (i.e., does not depend on our choice of $v_1$ above) because $\ker \alpha = \ker \beta$. Suppose that both $v_1,v_1' \in V_1$ satisfy $\beta(v_1)=\beta(v_1') = v_2$. We want to check that $\alpha(v_1) = \alpha(v_1')$. But since $v_1-v_1' \in \ker \beta$, $v_1-v_1' \in \ker \alpha$ and it follows that $\alpha(v_1-v_1') = 0$. Actually we only need $\ker \beta \subseteq \ker \alpha$ here.
Linearity follows from the linearity of $\alpha$ and $\beta$, so I won't elaborate on this part (unless you ask me to, of course).
A: There is already an excellent, clear, elementary answer here, but I just thought I would add another, slightly more abstract perspective.
Note that by the first isomorphism theorem, we have $V_1/\ker\alpha\cong \operatorname{im}\alpha$, and $V_1/\ker\beta\cong \operatorname{im}\beta=V_2$ since $\beta$ is onto. Since $\ker\alpha=\ker\beta$, this gives $V_2\cong\operatorname{im} \alpha$ via some map $\lambda: V_2 \to \operatorname{im}\alpha$, and checking the definitions of these maps, we have that $\lambda$ is precisely the map from Ivo's answer, and that $\alpha=\lambda\circ\beta$. Note that this shows automatically that the map is linear and well defined, although it certainly requires being comfortable with a less concrete construction.
