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Suppose that we have the following commutative diagram of groups and homomorphisms $$\newcommand\twoheaduparrow{\mathrel{\rotatebox{90}{$\twoheadrightarrow$}}} \begin{array} A & A_3 & {\hookrightarrow} & A_2 &{\twoheadrightarrow} & A_1 & \ \\ & \downarrow{} & &\downarrow{}& &\downarrow{}\\ & B_3 & \stackrel{}{\hookrightarrow} &B_2 & \stackrel{}{\twoheadrightarrow} & B_1 & & \\ & \downarrow{} & &\downarrow{}& &\downarrow{}\\ &C_3 & &C_2 & \stackrel{}{\twoheadrightarrow} & C_1 & & \end{array}$$ where all columns and rows are short exact sequences except the $3$rd row, where the map $C_2\to C_1$ is onto. (Sorry but not sure how to draw vertical arrows to denote $1$-$1$ maps and onto maps.)

Question

Is there a induced map $f:C_3\to C_2$ making the $3$rd row also a short exact sequence? Or at least can we prove that $f:C_3\to C_2$ is one-to-one?

I tried the five lemma but it does not seem to work here.

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  • $\begingroup$ The snake lemma will help here. $\endgroup$ – Hanno Jan 5 '15 at 15:28
  • $\begingroup$ Thanks @Hanno! So the answer to my question should be "yes"? $\endgroup$ – Zuriel Jan 5 '15 at 15:32
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Yes, there is such a map. This can be shown by diagram chase:

  1. For $c \in C_3$ choose $b \in b_3$ s.t. $b=(B_3\to C_3)(b)$. Then $f(c) := (B_2 \to C_2)\circ (B_3 \to B_2)(b)$ is well-defined.
  2. $f(c) \in \ker (C_2 \to C_1)$ follows from $(C_2 \to C_1)\circ (B_2 \to C_2) \circ (B_3 \to B_2) = (B_1 \to C_1)\circ (B_2 \to B_1) \circ (B_3 \to B_2)$ since the last composition is zero.
  3. To show $\operatorname{im}(f) = \ker(C_2 \to C_1)$, let $c_2 \in \ker(C_2 \to C_1)$ and choose $b_2\in B_2$ with $c_2=(B_2 \to C_2)(b_2)$. Hence $(B_2 \to B_1)(b_2) \in \ker (B_1 \to C_1) = \operatorname{im}(A_1 \to B_1)$. Hence we find $a_2 \in A_2$ s.t. $(B_2 \to B_1)(b_2)=(A_1 \to B_1) \circ (A_2 \to A_1)(a_2) = (B_2 \to B_1) \circ (A_2 \to B_2)(a_2)$. Hence $b_2 - (A_2 \to B_2)(a_2) \in \ker(B_2 \to B_1)=\operatorname{im}(B_3 \to B_2)$. Choose $b_3 \in B_3$ with $(B_3\to B_2)(b_3)=b_2 - (A_2 \to B_2)(a_2)$ and set $c_3 := (B_3 \to C_3)(b_3)$. Then $f(c_3)=(B_2 \to C_2)(b_2)=c_2$.
  4. The injectivity of $f$ can be shown in the same pattern as in 3. Hint: You have to use the injectivity of $A_1 \to B_1$ and of $B_3 \to B_2$.

Note that $f$ also makes the diagram commutative.

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  • $\begingroup$ Why was my answer downvoted ? Please give a reason. Thanks. $\endgroup$ – tj_ Jan 6 '15 at 22:22
  • $\begingroup$ Not sure who did it; your answer seems perfect to me. $\endgroup$ – Zuriel Jan 7 '15 at 17:23

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