Homology and right exact functors

qwirk · **Joined:** Fri, 3 Sep 2010 09:29:45 UTC **Posts:** 140

Let $F:\mathscr{A} \to \mathscr{B}$ be a covariant functor of abelian categories and $C^\bullet$ a chain complex in $\mathscr{A}$

I want to show that if

is right exact there is a natural map $FH^* \to H^*F$

There is some useful hints for the problem. First start with the exact sequence

$C^i \stackrel{d^i}{\to} C^{i+1} \to \text{coker}d^i \to 0$

Apply

to get

$F(C^i) \stackrel{F(d^i)}{\to} F(C^{i+1}) \to F(\text{coker}d^i) \to 0$

and then it is not hard to see $\displaystyle F(\text{coker}d^i) = \text{coker}F(d^i)$

We then use the exact sequence

$\text{im}d^i \to C^{i+1} \to \text{coker}d^i \to 0$
Apply F
$F(\text{im}d^i) \to F(C^{i+1}) \to F(\text{coker}d^i) \to 0$
$F(\text{im}d^i) \to F(C^{i+1}) \to \text{coker}F(d^i) \to 0$
Now $\displaystyle \text{coker}F(d^i) \simeq \frac{F(C^{i+1})}{\text{im}F(d^i)}$

Assuming this is correct thus far, how does this give an epimorphism $F \text{im}d^i \to \text{im} F(d^i)$ ?

Shadow · **Posted:** Thu, 12 Jan 2012 22:35:54 UTC

qwirk wrote:

Let $F:\mathscr{A} \to \mathscr{B}$ be a covariant functor of abelian categories and $C^\bullet$ a chain complex in $\mathscr{A}$

I want to show that if

is right exact there is a natural map $FH^* \to H^*F$

There is some useful hints for the problem. First start with the exact sequence

$C^i \stackrel{d^i}{\to} C^{i+1} \to \text{coker}d^i \to 0$

Apply

to get

$F(C^i) \stackrel{F(d^i)}{\to} F(C^{i+1}) \to F(\text{coker}d^i) \to 0$

and then it is not hard to see $\displaystyle F(\text{coker}d^i) = \text{coker}F(d^i)$

We then use the exact sequence

$\text{im}d^i \to C^{i+1} \to \text{coker}d^i \to 0$
Apply F
$F(\text{im}d^i) \to F(C^{i+1}) \to F(\text{coker}d^i) \to 0$
$F(\text{im}d^i) \to F(C^{i+1}) \to \text{coker}F(d^i) \to 0$
Now $\displaystyle \text{coker}F(d^i) \simeq \frac{F(C^{i+1})}{\text{im}F(d^i)}$

Assuming this is correct thus far, how does this give an epimorphism $F \text{im}d^i \to \text{im} F(d^i)$ ?

Do you mean homology or cohomology, your topic says one and your notation indicates the other.

qwirk · **Joined:** Fri, 3 Sep 2010 09:29:45 UTC **Posts:** 140

Cohomology - but what exactly is looking like homology?

Shadow · **Posted:** Fri, 13 Jan 2012 06:12:20 UTC

qwirk wrote:

Cohomology - but what exactly is looking like homology?

Reread what you posted as the topic.

outermeasure · **Posted:** Fri, 13 Jan 2012 13:23:55 UTC

qwirk wrote:

Let $F:\mathscr{A} \to \mathscr{B}$ be a covariant functor of abelian categories and $C^\bullet$ a chain complex in $\mathscr{A}$

I want to show that if

is right exact there is a natural map $FH^* \to H^*F$

There is some useful hints for the problem. First start with the exact sequence

$C^i \stackrel{d^i}{\to} C^{i+1} \to \text{coker}d^i \to 0$

Apply

to get

$F(C^i) \stackrel{F(d^i)}{\to} F(C^{i+1}) \to F(\text{coker}d^i) \to 0$

and then it is not hard to see $\displaystyle F(\text{coker}d^i) = \text{coker}F(d^i)$

We then use the exact sequence

$\text{im}d^i \to C^{i+1} \to \text{coker}d^i \to 0$
Apply F
$F(\text{im}d^i) \to F(C^{i+1}) \to F(\text{coker}d^i) \to 0$
$F(\text{im}d^i) \to F(C^{i+1}) \to \text{coker}F(d^i) \to 0$
Now $\displaystyle \text{coker}F(d^i) \simeq \frac{F(C^{i+1})}{\text{im}F(d^i)}$

Assuming this is correct thus far, how does this give an epimorphism $F \text{im}d^i \to \text{im} F(d^i)$ ?

Recall, for any morphism $f\colon A\to B$ in an abelian category, you have the 4 associated objects $\mathop{\mathrm{ker}}f,\mathop{\mathrm{coker}}f,\mathop{\mathrm{im}}f,\mathop{\mathrm{coim}}f$ and a canonical isomorphism $\mathop{\mathrm{coim}}f\xrightarrow[\tilde{f}]{\cong}\mathop{\mathrm{im}}f$ which gives the commutative diagram
$\begin{array}{ccc} 0&&0\\ \downarrow &&\uparrow\\ \mathop{\mathrm{ker}}f && \mathop{\mathrm{coker}}f\\ \downarrow &&\uparrow\\ A&\xrightarrow{\quad f\quad} & B\\ \downarrow &&\uparrow\\ \mathop{\mathrm{coim}}f &\xrightarrow[\tilde{f}]{\quad\cong\quad}&\mathop{\mathrm{im}}f\\ \downarrow &&\uparrow\\ 0&&0 \end{array}$
with exact columns.

Now, $\mathop{\mathrm{im}}f$ is the kernel of $B\to\mathop{\mathrm{coker}}f$ , so by the universal property of kernel, the map $F\mathop{\mathrm{im}}d^i\to FC^{i+1}$ factors through $\mathop{\mathrm{im}}Fd^i$ , and that is your morphism $F\mathop{\mathrm{im}}d^i\to\mathop{\mathrm{im}}Fd^i$ . And it is not hard to show it is an epimorphism (use $\mathop{\mathrm{coim}}Fd^i\to F\mathop{\mathrm{coim}}d^i$ together with the isomorphisms)

qwirk · **Joined:** Fri, 3 Sep 2010 09:29:45 UTC **Posts:** 140

Very nice!

Thank you for that

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