Consider the following commutative diagram, where rows are exact: $$\require{AMScd} \begin{CD} M_1 @>f_1>> M_2 @>f_2>> M_3 @>f_3>> M_4 @>f_4>> M_5 \\ @Vh_1VV @Vh_2VV @Vh_3VV @Vh_4VV @Vh_5VV \\ N_1 @>g_1>> N_2 @>g_2>> N_3 @>g_3>> N_4 @>g_4>> N_5 \end{CD} $$

All modules and homomorphisms above. Then the exercise asks me to check three things:

(a) If $h_1$ is an epimorphism and $h_4$ is a monomorphism, then $\ker h_3 = f_2(\ker h_2)$;

(b) If $h_2$ is an epimorphism and $h_5$ is a monomorphism, then $g_3^{-1}({\rm im}\,h_4) = {\rm im}\, h_3$;

(c) If $h_1, h_2, h_4$ and $h_5$ are isomorphisms, so is $h_3$.

I managed to solve item (b), and assuming item (a) too, I managed to solve item (c). I'm having trouble with item (a). The inclusion $\ker h_3 \supseteq f_2(\ker h_2)$ is easy to check.

Then take $x \in \ker h_3$. So $h_3(x) =0$, and $g_3(h_3(x)) = h_4(f_3(x)) = 0$. From this, $f_3(x)= 0$ and $x \in \ker f_3 = {\rm im}\,f_2$, so $x=f_2(y)$ for some $y$. If I could prove that $y\in \ker f_2$, I would be done.

What I can see is that $h_3(f_2(y)) = g_2(h_2(y)) =0$, so $h_2(y) \in \ker g_2 = {\rm im}\,g_1$, so $h_2(y)=g_1(z)$ for some $z$. But $z=h_1(w)$ for some $w$, so $h_2(y) = g_1(h_1(w))=h_2(f_1(w))$, and from here $y-f_1(w)\in \ker h_2$. But I can't prove that $f_1(w)\in\ker h_2$ to conclude the $y\in \ker h_2$ that I want.


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    $\begingroup$ Your diagram is quite nice. I just asked myself how to do this diagram in ;MSE. By the way, in Rotman's book on group theory, the $5$-lemma with10 groups and $13$ morphisms is proved just the way you started. So have a look there. Or see the references here. $\endgroup$ – Dietrich Burde Aug 13 '15 at 19:06
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    $\begingroup$ Sounds like something fitting the tag of [diagram-chasing], no? $\endgroup$ – Asaf Karagila Aug 13 '15 at 19:13
  • $\begingroup$ I didn't knew that such a tag existed. Seems relevant, thanks $\endgroup$ – Ivo Terek Aug 13 '15 at 19:14
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    $\begingroup$ @IvoTerek You may want to look at the source of my edit; try texdoc amscd for more information. Of course, in a LaTeX document, tikz-cd or Xy-pic would be preferable. $\endgroup$ – egreg Aug 13 '15 at 20:26
  • $\begingroup$ @egreg thanks for the edit. I didn't answered earlier because your code wasn't compiling, but I was browsing the site by mobile (a tablet, actually). I'm on a computer now and it works. I wonder what's the problem. Maybe I'll ask on meta about it. $\endgroup$ – Ivo Terek Aug 14 '15 at 18:57

You actually don't want $y \in \ker h_2$.

You have $f_2(y - f_1(w)) = f_2(y) = x$, and $$h_2(y-f_1(w)) = h_2(y) - h_2(f_1(w)) = h_2(y) - g_1(h_1(w)) = h_2(y) -g_1(z) = 0.$$

Exactly what we need.

  • $\begingroup$ I was pretty close then! Thanks :) $\endgroup$ – Ivo Terek Aug 13 '15 at 19:39

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