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Suppose that the following diagram of $R$-modules and $R$-homomorphisms is commutative and has exact rows. Show that, If $\alpha, \gamma$, and $g$ are onto, then so is $\beta$. $$\begin{array} KK & \stackrel{f}{\longrightarrow} & M & \stackrel{g}{\longrightarrow} & L\\ \downarrow{\alpha} & & \downarrow{\beta} & & \downarrow{\gamma}\\ K' & \stackrel{f'}{\longrightarrow} & M'& \stackrel{g'}{\longrightarrow} & L' \end{array} $$

So, I need to show that for every $m'\in M'$ there exists a $m\in M$ such that $\beta(m)=m'$. And we have the following maps which are onto.

$$\begin{array} KK & & M & \stackrel{g}{\longrightarrow} & L\\ \downarrow{\alpha} & & & & \downarrow{\gamma}\\ K' & & M'& & L' \end{array} $$

Suppose $g'(m')=l'\tag1$ then using the surjectivity of $g,\gamma$ we can say that $$\exists l\in L:\gamma(l)=l'\\\exists m\in M:g(m)=l$$ And from commutativity, \begin{align} \gamma g(m)&=g'\beta (m)\\ \gamma(l) & =g'\beta (m)\\ l' &= g'\beta (m) \end{align} Does it imply $\beta(m)=m'$ using (1)? I didn't use exactness of the rows. Is my approach correct? If not, then please suggest how to mentally picture to solve this kind of problem.

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  • $\begingroup$ I guess to use $(1)$ to argue $\beta(m)=m'$, I need to prove injectivity of $g'$ first $\endgroup$
    – N00BMaster
    Sep 13, 2023 at 12:03
  • $\begingroup$ you just proved that $g'$ is surjective, you will need to use exactness and surjectivity of $\alpha$ in order to control the difference of elements in the same preimage under $g'$. Short hint: write $m'=k+\beta \left(m\right)$ for $k \in \mathrm{ker} f'$ $\endgroup$
    – Felix
    Sep 13, 2023 at 12:51
  • $\begingroup$ I am afraid that I couldn't complete that using your short hint. Can you elaborate it? @Enkidu $\endgroup$
    – N00BMaster
    Sep 13, 2023 at 13:48
  • $\begingroup$ essentially, you already got a preimage of $\beta (m)$ so you need to find a preimage of $k$ and as this is in $\mathrm{ker} g' = \mathrm{im} (f')$ ... btw. sorry for the typo, meant $k\in \mathrm{ker}(g') of course$ $\endgroup$
    – Felix
    Sep 13, 2023 at 14:47
  • $\begingroup$ I suspect I need to use surjectivity to get an element from $K$ using $\text{im}\:(f')=\text{ker}\:g'$ and then apply commutativity of $\alpha f'$ and $\beta f$. But I didn't see where we are going? It seems like I can't imagine the broader picture for this theorem. @Enkidu $\endgroup$
    – N00BMaster
    Sep 13, 2023 at 15:21

2 Answers 2

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The start of your argument is a bit confusing, due to the introduction of $\ell'$.Allow me first to rewrite slightly your arguments.

Let's start with an $m'\in M'$. We'll prove that it lies in the image of $\beta$.

We have $g'(m')\in L'$, and by surjectivity of $\gamma$, there is an $\ell\in L$ such that $g'(m')=\gamma(\ell)$.

Now you have to use commutativity of the diagram. Using surjectivity of $g$, we have $\ell=g(m)$ for some $m\in M$, so that $g'(m')=\gamma g(m)=g'\beta(m)$. So far, this is what you obtained.

Now we have $g'(m'-\beta(m))=0$, so $m'-\beta(m)\in \ker(g')=im(f')$ by exactness of the bottom row. Thus $m'-\beta(m)=f'(k')$ for some $k'\in K'$. By surjectivity of $\alpha$, we get $k\in K$ such that $m'-\beta(m)=f'\alpha(k)$, and by commutativity, $m'-\beta(m)=\beta f(k)$, that is $m'=\beta(m+f(k))$.

Conclusion. The ideas you didn't want to pursue because you didn't know where this was going were exactly the right ideas.

Sometimes, you just have to accept that the solutions of some exercises are just mechanical and that there is no deep meaning. This is typical of these "diagram chasing" exercises. The solution of such exercices is 100% of the time: apply commutativity/surjectivity/injectivity/ exactness in various orders until you get what you want...

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  • $\begingroup$ Thanks @GreginGre for your answer. $\endgroup$
    – N00BMaster
    Sep 14, 2023 at 20:13
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MacLane's "Homology" has the following variant of five-lemma.

Consider commutative diagram with exact rows. If $ \alpha,\gamma$ are epimorphisms and $\delta$ is monomorphism then $\beta$ is epimorphism: $$\begin{array} KK & \stackrel{f}{\longrightarrow} & M & \stackrel{g}{\longrightarrow} & L & \stackrel{h}{\longrightarrow} & N \\ \downarrow{\alpha} & & \downarrow{\beta} & & \downarrow{\gamma} & & \downarrow{\delta}\\ K' & \stackrel{f'}{\longrightarrow} & M'& \stackrel{g'}{\longrightarrow} & L' & \stackrel{h'}{\longrightarrow} & N' \end{array} $$

In your case $N=0$ so $\delta$ is monomorphism.

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