Cohomology-exact sequences-1

A sequence of homomorphisms of vector spaces

$A \xrightarrow{f(x)} B \xrightarrow{g(x)} C$

is called "exact sequence" if $im f = ker g$.

One way of thinking about is as follows. Assume that $f(x)=g(x)=d$. Assume that $d$ applied to any element "dirties" it and second application $d$ to "dirtied" element sends it to $0$. Thus image of $d$ or $f$ is "dirtied" and because $g$  or second $d$ sends all such "dirtied" elements to zero, clearly $im\; d (or f)  = ker\; d( or g)$.

Short exact sequence has the form $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$.

Note for a sequence to be exact all the terms except for the first and last need to be exact.

The sequence $0 \xrightarrow{f} A \xrightarrow{g} B$ exact means $im\;f=ker;g=0$. Since only element in $ker\;g$ is $0$, $g$ is injective.

Similarly, in case $A \xrightarrow{f} B \xrightarrow{g} 0$, then all of $B$ is $ker\;g$. Hence, $im\;f=ker\;g=B$ and $f$ is surjective.

Problem 24.1 (Tu - An Introduction to Manifolds)

Given an exact sequence,
$A \xrightarrow{f(x)} B \xrightarrow{g(x)} C$

Show that $f$ is surjective iff $g$ is zero map.

Prf:

Assume $g$ is zero map. Then $ker\;g=B$. Using this in the definition of exact sequence results in $im\;f=ker\;g=B$. Hence $f$ is surjective.

Assume $f$ is surjective. From exact sequence definition, this means $im\;f=ker\;g=B$. $ker\;g=B$ implies $g$ is a zero map.

Show that $f$ is zero map iff $g$ is injective.

Prf:

Assume $g$ is injective. This means $ker\;g=0$. Using definition of exact sequence, results in $im\;f=ker\;g=0$. This $f$ is zero map.

Assume $f$ is zero map which means $im\;f=0$. Exact sequence implies $ker\;g=0$.Hence, $g$ is injective.

Problem 24.2. Four term exact sequence.

Four term sequence of vector spaces $0 \rightarrow A \xrightarrow{f} B \rightarrow 0$ is exact iff $f: A \rightarrow B$ is an isomorphism.

Prf:

Assume $f$ is an isomorphism which means $f$ is an injective and surjective function. $f$ is injective means $ker\;f=0$. And this is clearly image of previous zero map. $f$ is surjective and subsequent map is zero map. Then $im\;f$ is same as kernel of subsequent map.Both the conclusions yield an exact sequence.

Assume sequence is exact. Then image of zero map which is zero is clearly in kernel of $f$. Image of $f$ is same as kernel of subsequent (zero map). Hence $f$ is surjective. Thus $f$ is an isomorphism.

If $A \xrightarrow{f} B \rightarrow C \rightarrow 0$ is exact, then $C=coker\;f = \frac{B}{im\;f}$.

Rewrite sequence as

$A \xrightarrow{f} B \xrightarrow{g} C \rightarrow 0$

Given the sequence is exact, $im\;f=ker\;g$. Cokernel of $f$ is defined as $B/im\;f$. This can be written as $B/ker\;g$. First isomorphism theorem yields $B/ker\;g \cong im \;g$. However $im\;g = ker(C\rightarrow 0)$ which is all of $C$. Hence,
$B/ker\;g \cong C$.