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 $\gamma$.

  • $\begingroup$ 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. $\endgroup$
    – DonAntonio
    Oct 3 '16 at 19:05
  • 1
    $\begingroup$ @DonAntonio, why does it matter? $\endgroup$
    – Hubble
    Oct 3 '16 at 19:06
  • 1
    $\begingroup$ 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. $\endgroup$ Oct 3 '16 at 19:13
  • 1
    $\begingroup$ @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 $\endgroup$ Oct 3 '16 at 19:15
  • 1
    $\begingroup$ 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 $C \to C/\delta \to A/\ker\alpha \to B$. $\endgroup$ 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 $a \in A$, $\gamma(\beta(a)) = \alpha(a)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.