# A Sequence induced by short exact sequence

1. Dec 31, 2017

### PsychonautQQ

This isn't homework, it's a proof left to the reader as I self study Munkre's 'Elements of Algebraic Topology'

Prove that if the sequence
$A_1 --> A_2 --> A_3 --> A_4 --> A_5$ is exact

Then so is the induced sequence:

$0 --> cok(a_1) --> A_3 --> ker(a_4) --> 0$

where $a_1$ and $a_4$ are the maps from $A_1 --> A_2$ and $A_4 --> A_5$ respectively.
Also, let $b_i$ represent the map from the i'th group in the induced sequence to the (i+1)'th

First of all, what exactly is an induced sequence? Secondly, I started trying to figure out the proof and failed. Here are Some of my thoughts:

If $x,y \in cok(a_1)$ and $b_2(x)=b_2(y)$, wwts $x=y$, because that will mean that $b_2$ is injective, which it must be if it's going to be exact because $im(b_1) = 0 = ker(b_2)$.

Since $x,y \in cok(a_1)$, there are no elements of $A_1$ that map to them in $A_2$.

And ahhh yeah actually I'm just confused... anyone got any tips for this proof?
This means there are no elements of $A_1$ that map to $x,y$ in $A_2$

2. Dec 31, 2017

### Staff: Mentor

Exactness of the first sequence means especially $\operatorname{im} a_1 \subseteq \operatorname{ker}a_2$ and $\operatorname{im}a_3\subseteq \operatorname{ker}a_4$. Therefore we get induced mappings
$$\overline{a}_2\, : \, A_2/\operatorname{im}a_1 \longrightarrow A_3 \quad \text{ and } \quad \overline{a}_3 \, : \, A_3 \longrightarrow \operatorname{ker} a_4 \subseteq A_4$$
As the image of $a_1$ is mapped to the kernel of $a_2$, $\overline{a}_2$ is well-defined, as two representatives are mapped on the same element. This is meant by induced. The second mapping $\overline{a}_3$ is simply the restriction on a smaller codomain. Since $a_3$ hits only elements in the kernel of $a_4$ anyway, there is nothing else needed to be shown, we already have a mapping on $\operatorname{ker}a_4$.
You have to show that $\overline{a}_2$ is injective and $\overline{a}_3$ is surjective. That is $\overline{a}_2(\overline{x}) = \overline{a}_2(\overline{y})$ implies $x \in y \circ \operatorname{im}a_1$ whatever your operation $\circ$ is, probably addition.

And every element in $\operatorname{ker}a_4$ is actually hit by $\overline{a}_3$, i.e. for every $y \in A_4$ with $y \in \operatorname{ker}a_4 = \{e\}$ - whatever your neutral element $e$ is, probably $e=0$ - there is an $x \in A_3$ with $a_3(x)=y$.

The second case is probably a bit easier than the first and we will also probably have to use exactness at $A_3$, which we haven't until now, i.e. $\operatorname{im}a_2 \subseteq \operatorname{ker}a_3\,$.

3. Dec 31, 2017

### WWGD

I think that, in a literal sense, induced maps arise from the standard isomorphism theorems in Algebra.

4. Jan 1, 2018

### mathwonk

i would quibble with the way this problem is phrased. I.e. one has to show both that there IS an induced sequence and also that it is exact. This may be part of your confusion. I.e. a map A2-->A3 always induces a map A2/ker(a2) --> A3, but one needs exactness, or at least that this is a complex, to know that the image of the previous map is contained in the kernel, so that there is an induced map on cok(a1) = A2/im(A1). the same holds at the other end to know that im(a3) is contained in ker(a4). so if so, i fault the author for sloppy language, although this author is seldom accused of that.