Prove that the map is injective if the rows and the columns are exact in the commutative diagram (self-ansered).

Question

This is an exercise I found a little painful, so I choose to archive it here.

Consider the following diagram of $R$-modules with exact rows and columns: $$\begin{array} A & && && & & & & &&0 & \ \\ & && & & & && &&&\downarrow{}\\ & && && & & B &\stackrel{g}{\longrightarrow}& C & \stackrel{h}{\longrightarrow}&D \ \\ & && && & &\downarrow{\beta}& &\downarrow{\gamma}&&\downarrow{\delta}\\ & && && A' & \stackrel{f_1}{\longrightarrow} & B' &\stackrel{g_1}{\longrightarrow} & C' & \stackrel{h_1}{\longrightarrow}&D' \\ & && && \downarrow{\alpha_1} & &\downarrow{\beta_1}& &\downarrow{\gamma_1}&&\downarrow{\delta_1}\\ & 0 && {\longrightarrow} &&A'' & \stackrel{f_2}{\longrightarrow} &B'' & \stackrel{g_2}{\longrightarrow} & C'' & \stackrel{h_2}{\longrightarrow} & D'' & \\ & && && \downarrow{\alpha_2}& &\downarrow{\beta_2}& &\\ & && && A'''&\stackrel{f_3}{\longrightarrow} & B'''& & & & \\ &&&&&\downarrow{}&&\\ &&&&&0&& \end{array} $$ Show that $f_3$ is always injective.

Answer 1

Here is my own (silly) solution.

Pick $a'''\in A'''$ such that $f_3(a''')=0$. Since $\alpha_2$ is surjective, there exist $a''\in A''$ such that $\alpha_2(a'')=a'''$.

Case 1: Assume that $a''=0$, then clearly $a'''=\alpha_2(a'')=0$.

Case 2: Assume that $a''\neq0$, one sees that $b'':=f_2(a'')\neq0$ since $f_2$ is injective. Note that $$\beta_2(b'')=\beta_2(f_2(a''))=(\beta_2\circ f_2)(a'')=(f_3\circ\alpha_2)(a'')=0$$ So $b''\in\ker(\beta_2)=\text{Im}(\beta_1)$, hence there exists $b'\in B'$ such that $b''=\beta_1(b')$.

Case 2.1: Assume that $b'\in\ker(g_1)=\text{Im}(f_1)$, one then has some $a'\in A'$ such that $f_1(a')=b'$. Here $a'$ must be nonzero, otherwise $b'=0$ and then $0\neq b''=\beta_1(b')=\beta_1(0)=0$, a contradiction. Note that $$f_2(a'')=b''=\beta_1(b')=\beta_1(f_1(a'))=f_2(\alpha_1(a'))$$ one has $a''=\alpha_1(a')$ since $f_2$ is injective. Therefore, $a''\in\text{Im}(\alpha_1)=\ker(\alpha_2)$ and thus $a'''=\alpha_2(a'')=0$.

Case 2.2: Assume that $b'\notin\ker(g_1)$, then $c':=g_1(b')\neq0$ and thus $h_1(c')=0$. Also, one has $$\gamma_1(c')=\gamma_1(g_1(b'))=g_2(\beta_1(b'))=g_2(b'')=0$$ because $b''\in\text{Im}(f_2)=\ker(g_2)$. Then there exists nonzero $c\in C$ such that $\gamma(c)=c'$ and $h(c)=0$ (here we use the injectivity of $\delta$). It follows a nonzero $b\in B$ with $g(b)=c$. Let $b'_0:=\beta(b)\neq0$, we know that $$g_1(b')=g_1(b'_0)=c'$$ but $b'\neq b'_0$, otherwise $b'\in\ker(\beta_1)$ and $b''=0$, a contradiction. Let $m:=b'-b'_0$, from $g_1(m)=c'-c'=0$, we know that there is some nonzero $n\in A'$ such that $f_1(n)=m$. Let $k=\alpha_1(n)$, note that $$f_2(k)=\beta_1(f_1(n))=\beta_1(m)=\beta_1(b'-b'_0)=b''-0=b''$$ it must be $k=a''$ because $f_2$ is injective, which implies that $a''=\alpha_1(n)\in\text{Im}(\alpha_1)=\ker(\alpha_2)$. Therefore, one still has $a'''=\alpha_2(a'')=0$.

In conclusion, one always has $a'''=0$ when $f_3(a''')=0$. It implies the statement.