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

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.

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.