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Donu Arapura
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I guess if you're asking a question like this, you'd likeLet me give a really simple mindedsomewhat elementary answer. Here's an attempt. LikeLike a lot of things, this stuff (presumably) came out of calculus. Start with integrals of the form $\int_C \omega$, where $C$ is closed curve and $\omega$ a differential. If one assumes that $d\omega=0$, then by Stokes the integral depends only the homology class of $C$ i.e. $C$ can be replaced by any $C'$ such that $C-C'$ is boundary of something. Dually the integral is unchanged if $\omega$ is replaced by $\omega+d(something)$. In other words, it depends only on the cohomology class of $\omega$.

This can be carried out in higher dimensions, as well. At first glance cohomology seems completely dual to homology, and therefore seemingly redundant. But in fact it has more structure. Since you multiply (wedge) differential forms together, cohomology becomes a ring. This This is still true in more general settingsapproaches such as singular cohomology. On the homology side, one has an intersection pairing, but this is harder to describe and only available for really "nice" spaces.

Perhaps another feature of cohomology worth mentioning is that is contravariant: cohomology classes pullback from the target to the source under a map of spaces. This important in the theory of characteristic classes, where such classes are pulled back from to maps to certain universal spaces. Such classes measure the amount of "twisting" of bundles.

I guess if you're asking a question like this, you'd like a really simple minded answer. Here's an attempt. Like a lot of things, this stuff (presumably) came out of calculus. Start with integrals of the form $\int_C \omega$, where $C$ is closed curve and $\omega$ a differential. If one assumes that $d\omega=0$, then by Stokes the integral depends only the homology class of $C$ i.e. $C$ can be replaced by any $C'$ such that $C-C'$ is boundary of something. Dually the integral is unchanged if $\omega$ is replaced by $\omega+d(something)$. In other words, it depends only on the cohomology class of $\omega$.

This can be carried out in higher dimensions, as well. At first glance cohomology seems completely dual to homology, and therefore seemingly redundant. But in fact it has more structure. Since you multiply (wedge) differential forms together, cohomology becomes a ring. This is still true in more general settings. On the homology side, one has an intersection pairing, but this is harder to describe and only available for really "nice" spaces.

Let me give a somewhat elementary answer. Like a lot of things, this stuff (presumably) came out of calculus. Start with integrals of the form $\int_C \omega$, where $C$ is closed curve and $\omega$ a differential. If one assumes that $d\omega=0$, then by Stokes the integral depends only the homology class of $C$ i.e. $C$ can be replaced by any $C'$ such that $C-C'$ is boundary of something. Dually the integral is unchanged if $\omega$ is replaced by $\omega+d(something)$. In other words, it depends only on the cohomology class of $\omega$.

This can be carried out in higher dimensions, as well. At first glance cohomology seems completely dual to homology, and therefore seemingly redundant. But in fact it has more structure. Since you multiply (wedge) differential forms together, cohomology becomes a ring. This is still true in more general approaches such as singular cohomology. On the homology side, one has an intersection pairing, but this is harder to describe and only available for really "nice" spaces.

Perhaps another feature of cohomology worth mentioning is that is contravariant: cohomology classes pullback from the target to the source under a map of spaces. This important in the theory of characteristic classes, where such classes are pulled back from to maps to certain universal spaces. Such classes measure the amount of "twisting" of bundles.

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Donu Arapura
  • 35.2k
  • 2
  • 94
  • 160

I guess if you're asking a question like this, you'd like a really simple minded answer. Here's an attempt. Like a lot of things, this stuff (presumably) came out of calculus. Start with integrals of the form $\int_C \omega$, where $C$ is closed curve and $\omega$ a differential. If one assumes that $d\omega=0$, then by Stokes the integral depends only the homology class of $C$ i.e. $C$ can be replaced by any $C'$ such that $C-C'$ is boundary of something. Dually the integral is unchanged if $\omega$ is replaced by $\omega+d(something)$. In other words, it depends only on the cohomology class of $\omega$.

This can be carried out in higher dimensions, as well. At first glance cohomology seems completely dual to homology, and therefore seemingly redundant. But in fact it has more structure. Since you multiply (wedge) differential forms together, cohomology becomes a ring. This is still true in more general settings. On the homology side, one has an intersection pairing, but this is harder to describe and only available for really "nice" spaces.