Splicing together two short exact sequences

Question

This question is from my module theory assignment and I need help in solving the (a) only.

(a) If $0\to A\to B\xrightarrow{f} C\to 0$ and $0\to C\xrightarrow{g} D\to E\to 0$ are short exact sequences of modules, then the sequence $0\to A\to B\xrightarrow{gf} D\to E\to 0$ is exact.
(b) Show that every exact sequence my be obtained by splicing together suitable short exact sequences as in (a).

Show that isomophism of short exact sequences is an equivalence relation.

$1$st diagram means that $C=\operatorname{im}(f)$ and $2$nd diagram means that $\ker(g)=0$. But I am not able to understand what it means by the $3$rd diagram, i.e. how to interpret it? Usually in case of exact, at least $2$ functions are given simultaneously so that equality can be written in terms of image of one and kernel of other. But in this case I am not sure how to interpret as only one diagram is given.

Thanks!

Answer 1

The items (a) and (b) are exercise 13 on page 180 of Hungerford's Algebra book.

item (a): Suppose $0\to A\xrightarrow{h} B\xrightarrow{f} C\to 0$ and $0\to C\xrightarrow{g} D\xrightarrow{k} E\to 0$ are short exact sequences. Then we have:

1. $h$ and $g$ are monomorphisms,
2. $f$ and $k$ are epimorphisms,
3. $\mathrm{Im}(h)= \mathrm{Ker}(f)$ and
4. $\mathrm{Im}(g)= \mathrm{Ker}(k)$

Since $f:B \to C$ and $g:C \to D$ are homomorphisms, we have that $gf:B \to D$ is an homomorphism. Moreover, since $g$ is a monomorphism, $\mathrm{Ker}(gf)= \mathrm{Ker}(f)$, and, since $f$ is an epimorphism, $\mathrm{Im}(gf)= \mathrm{Im}(g)$. So we have:

1. $h$ is a monomorphism,
2. $gf$ is an homomorphism
3. $k$ is an epimorphisms,
4. $\mathrm{Im}(h)= \mathrm{Ker}(gf)$ and
5. $\mathrm{Im}(gf)= \mathrm{Ker}(k)$

So, $0\to A \xrightarrow{h} B\xrightarrow{gf} D\xrightarrow{k} E\to 0$

item (b): It is a direct consequence that each step $B \xrightarrow{\phi} D$ in a exact can be obtained by slicing together the following short exact sequences: $$0\to \mathrm{Ker}(\phi)\xrightarrow{h} B\xrightarrow{f} B/\mathrm{Ker}(\phi)\to 0$$
and $$0\to B/\mathrm{Ker}(\phi)\xrightarrow{g} D\xrightarrow{k} D/\mathrm{Im}(\phi)\to 0$$ where $h$ is the inclusion of $\mathrm{Ker}(\phi)$ into $B$, $f$ is the canonical epimorphism from $B$ onto $B/\mathrm{Ker}(\phi)$, $g$ is quotient monomorphism from $B/\mathrm{Ker}(\phi)$ into $D$ (defined by $g(\overline{x})= \phi(x)$) and $k$ is is the canonical epimorphism from $D$ onto $D/\mathrm{Im}(\phi)$.

Remark: The "last line" of the question above is "Show that isomophism of short exact sequences is an equivalence relation". It is not related to the previous two items and it is an easy consequence of the definition of isomophism of short exact sequences and the definition of equivalence relation.

Answer 2

From what I understand, you are only used to the concept of short exact sequences, i.e. sequences of the form

$$ 0\to A\to B\to C\to 0\,, $$

missing the general concept of an exact sequence. The latter is a sequence $(A_i,d_i)$ of modules $A_i$ and homomorphisms $d_i\colon A_i\to A_{i+1}$ (conventions vary in which direction the maps go) such that $\ker(d_{i+1})=\operatorname{im}(d_i)$ for all $i$, which ones refers to as exactness at $A_i$ as the $\ker(d_{i+1}),\operatorname{im}(d_i)\le A_i$. As I do not intend to talk about this in more generality, we may as well assume that this sequence is in fact of the form

$$ 0=A_0\xrightarrow{d_0} A_1\xrightarrow{d_1} A_2\to\cdots\to A_{n-1}\xrightarrow{d_{n-1}} A_n\xrightarrow{d_n} A_{n+1}=0\,. $$

In this case, $\ker(d_{i+1})=\operatorname{im}(d_i)$ dictates that, in particular,

$$ \ker(d_1)=\operatorname{im}(d_0)=0\quad\text{and}\quad\operatorname{im}(d_{n-1})=\ker(d_n)=A_n $$

which includes your observations as special case.

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Now suppose we are given two short exact sequences

$$ 0\to A\xrightarrow{\iota} B\xrightarrow{f} C\to 0\quad\text{and}\quad0\to C\xrightarrow{g} D\xrightarrow{\pi} E\to 0 $$

and we want to show that the induced sequence

$$ 0\to A\xrightarrow{\iota} B\xrightarrow{gf} D\xrightarrow{\pi} E\to 0 $$

is then exact too. Exactness at $A$ and $E$ is still given as we use the same maps. So the relevant things to show are exactness at $B$ and at $D$.

The fastest way of doing this is to realize that

$$ \ker(gf)=\ker(f)\quad\text{and}\quad\operatorname{coker}(gf)=\operatorname{coker}(g)\,, $$

which follows from $g$ being a monomorphism and $f$ being an epimorphism (which is equivalent to being injective and surjective but requires some work), and using an alternative characterization (proposition $2.4$) of exactness in terms of cokernels. However, this might be a bit over the top. The main idea stands, nonetheless, i.e. using that $g$ and $f$ are injective and surjective (i.e. mono- and epimorphisms), resp., which then elementwise translates to

\begin{align*} ((g\circ f)\circ\iota)(x)=0\,&\iff\,(f\circ\iota)(x)=0 \\ (\pi\circ(g\circ f))(x)=0\,&\iff\,(\pi\circ g)(x)=0 \end{align*}

from where it should be clear how to apply the exactness of the original sequences.

Can you take it from here? If some terminology or ideas are unclear (mono-/epimorphisms, cokernels, etc.) please let me know. I can try to make the answer more elementary or include an explanation.