Homomorphisms : If $\ker\beta \subseteq \ker\alpha$ and $\beta$ is onto then $\alpha = \gamma \circ \beta$.

Let $\alpha : A \to B$ and $\beta : A \to C$ be homomorphisms with $\beta$ onto and $\ker\beta \subseteq \ker\alpha$. Show there is a homomorphism $\gamma : C \to B$ such that $\alpha = \gamma \circ \beta$.

It seems that I need to use the first and second isomorphism theorems. By the first isomorphism theorem I get that:

By $\beta$ onto I get an isomorphism from $A/\ker\beta$ to $C$

and from the second isomorphism theorem I get that :

by $\ker\beta \subseteq \ker\alpha$ we get $A/\ker\alpha \simeq (A/\ker\beta)\Big / (\ker\alpha \, / \ker\beta)$

I am not sure how to piece these together to get this $\gamma$.

• Homomorphisms... of what ? Groups, rings, algebras, linear spaces,...? Very probably these are abelian groups and you're about to work with exact sequences, split short sequences and etc. Oct 3 '16 at 19:05
• @DonAntonio, why does it matter? Oct 3 '16 at 19:06
• From a universal property, you get a homomorphism $\overline{\alpha} \colon A/\ker \beta \to B$ such that $\alpha = \overline{\alpha} \circ \pi$, where $\pi \colon A \to A/\ker \beta$ is the canonical map. Oct 3 '16 at 19:13
• @DonAntonio It is universal algebra, all that matters is that $A,B,C$ are similar algebras, so doesn't really matter in the context of the problem what algebras they are Oct 3 '16 at 19:15
• Via the first isomorphism you obtain $A/\ker\alpha \simeq C/\delta$ with $\delta\in \mathop{Con} C$ corresponding to $\ker\alpha / \ker\beta$. But $A/\ker\alpha \simeq \alpha[A]$ canonically. So you can take a path like $C \to C/\delta \to A/\ker\alpha \to B$. Oct 4 '16 at 0:14

Rather than thinking, "Which theorems should I use to prove this?", I want to ask instead, "What could this map $\gamma:C \longrightarrow B$ possibly be?" To that end, suppose $c$ is in $C$. How can we map $c$ over to $B$? Well, the only way we know of getting anything into $B$ is the map $\alpha:A \longrightarrow B$, but $c$ isn't in $A$. Ah, but $\beta$ is surjective, so $c$ has a pre-image (at least one) in $A$. Alright, let's take $a$ to be a pre-image of $c$, so that $\beta(a) = c$. Maybe we can define $\gamma(c) = \alpha(a)$. Does this yield a well-defined map?
The main issue is this: If $c$ has more than one pre-image under $\beta$, might $\alpha$ send them to distinct points in $B$? If so, then we're sunk; at least, we'll have have to find a different approach. To find out, let $a$ and $a'$ both be pre-images of $c$. We need to show that $\alpha(a) = \alpha(a')$. But ${\rm ker}(\beta) \subset {\rm ker}(\alpha)$, which implies that $\alpha(x) = \alpha(y)$ whenever $\beta(x) = \beta(y)$. Since $\beta(a) = \beta(a')$, it follows that $\alpha(a) = \alpha(a')$, implying that our $\gamma$ is a well-defined function (of sets). And since $\alpha$ is a morphism, it follows that $\gamma$ is as well.
finally, does $\alpha = \gamma \circ \beta$? Yes, essentially by definition: Since $a$ is itself a pre-image of $\beta(a)$, our definition of $\gamma$ implies that, for all $a \in A$, $\gamma(\beta(a)) = \alpha(a)$.