Originally Posted by

**ModusPonens** Hello

I've done an exercise in which the 3rd question was the lemma of 5 itself. Now, I've done it in a non conventional way and my teacher says it's wrong but she didn't point out exactly what was wrong with my solutions. I'll just post one of them and please tell me where exactly it fails, if it indeed fails (I'm still convinced it doesn't fail):

$\displaystyle \begin{array}{ccccccccc} M_1&\stackrel{f_1}\longrightarrow& M_2&\stackrel{f_2}\longrightarrow& M_3&\stackrel{f_3}\longrightarrow& M_4&\stackrel{f_4}\longrightarrow& M_5\\ \hspace*{9pt}\downarrow\,\scriptstyle h_1&& \hspace*{9pt}\downarrow\,\scriptstyle h_2&& \hspace*{9pt}\downarrow\,\scriptstyle h_3&& \hspace*{9pt}\downarrow\,\scriptstyle h_4&& \hspace*{9pt}\downarrow\,\scriptstyle h_5\\ N_1&\stackrel{g_1}\longrightarrow& N_2&\stackrel{g_2}\longrightarrow& N_3&\stackrel{g_3}\longrightarrow& N_4&\stackrel{g_4}\longrightarrow& N_5 \end{array}$

Hhmm. Imagine the $\displaystyle h_i$ from $\displaystyle M_i$ to $\displaystyle N_i$. Now, the exercise is: Consider the above commutative diagram in which each horizontal sequence is exact. If $\displaystyle h_1$ is an epimorphism and $\displaystyle h_4$ is a monomorphism, then $\displaystyle Ker h_3=f_2(Ker h_2)$.

My proposal of solution doesn't need that $\displaystyle h_4$ is a monomorphism, and is done in the following way:

$\displaystyle f_2(Ker h_2)=f_2(h_2^{-1}(\{0\}))=f_2(f_1(h_1^{-1}(g_1^{-1}(\{0\}))))=\{0\}$

$\displaystyle h_3^{-1}(\{0\})=f_2(f_1(h_1^{-1}(g_1^{-1}(g_2^{-1}(\{0\})))))=\{0\}$

Where is this wrong, if at all?