3
$\begingroup$

Consider the following commutative diagram of homomorphisms of abelian groups $$\begin{array} 00&\stackrel{f_1}{\longrightarrow}&A& \stackrel{f_2}{\longrightarrow}&B& \stackrel{f_3}{\longrightarrow}&C&\stackrel{f_4}{\longrightarrow}& D &\stackrel{f_5}{\longrightarrow}&0\\ \downarrow{g_1}&&\downarrow{g_2}&&\downarrow{g_3}&&\downarrow{g_4}&&\downarrow{g_5}&&\downarrow{g_6}\\ 0&\stackrel{h_1}{\longrightarrow}&0& \stackrel{h_2}{\longrightarrow}&E& \stackrel{h_3}{\longrightarrow}&F&\stackrel{h_4}{\longrightarrow} &0 &\stackrel{h_5}{\longrightarrow}&0 \end{array} $$

Suppose the horizontal rows are exact ($\mathrm{ker}(f_{i+1})=\mathrm{Im}(f_i) $)

Suppose we know that $g_4:C\rightarrow F$ is an isomorphism.

How to deduce that $D=0$?

All what I could get is that $h_3:E\rightarrow F$ is an isomorphism and $f_4:C\rightarrow D$ is surjective.

$\endgroup$

1 Answer 1

Reset to default
7
$\begingroup$

This is wrong. Consider

\begin{array}{ccccccccccc} 0 & \to & 0 & \to & 0 & \to & A & \to & A & \to & 0\\ \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow\\ 0 & \to & 0 & \to & A & \to & A & \to & 0 & \to & 0 \end{array}

where all maps $A \to A$ are the identity.

$\endgroup$
7
  • $\begingroup$ @Theo Buehler : i see what you mean but if i show that $f_3$ is an isomorphism then i'm done. let me ask this question: consider the commmutative diagram of group homomorphisms $$\begin{matrix} A&\stackrel{f}{\rightarrow}&B\\ \downarrow{g}&&\downarrow{k}\\ C&\stackrel{h}{\rightarrow}&D \end{matrix} $$ if $h$ and $k$ are iso does this imply that $f$ is an iso? $\endgroup$
    – palio
    Jun 7, 2011 at 15:50
  • 3
    $\begingroup$ @palio: no. The middle square above, that is, \begin{array}{ccc} 0 & \to & A \\ \downarrow & & \downarrow \\ A & \to & A \end{array} shows that this isn't true. However, if you know that the square is a pull-back and $h$ is an iso then you can conclude that $f$ is an isomorphism. By the way: in your original diagram $D$ is zero if and only if $f_3$ is surjective. $\endgroup$
    – t.b.
    Jun 7, 2011 at 15:57
  • $\begingroup$ @palio: Could you please start accepting the answers that were helpful to you? This is done by clicking on the grey check mark sign on the left of the answers. There is a number 0% under your name showing that you haven't accepted any answers so far. People here will react much more favorably and helpfully if you have accepted at least some answers. This is how this site works. $\endgroup$
    – t.b.
    Jun 7, 2011 at 16:17
  • $\begingroup$ @ Theo Buehler : i didn't know about that!! i will do it of course!!! thanks. $\endgroup$
    – palio
    Jun 7, 2011 at 17:22
  • $\begingroup$ @palio: Glad to hear that! If I see correctly you clicked on the triangle next to my comment. What I meant is the grey √-sign on the left next to the very first thing I wrote :) $\endgroup$
    – t.b.
    Jun 7, 2011 at 17:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .