If A f⟶ B g⟶ C h⟶ D k⟶ E is exact, prove that there is an exact sequence 0 →coker f α⟶ C β⟶ ker

If A f⟶ B g⟶ C h⟶ D k⟶ E is exact, prove that there is an...

Key Concepts

Exact Sequence

An exact sequence is a sequence of morphisms between objects (such as groups, modules, or vector spaces) in which the image of one morphism equals the kernel of the subsequent morphism. This concept is fundamental in algebra and homological algebra as it captures the precise manner in which algebraic structures fit together and encodes information about injectivity and surjectivity of maps in the sequence.

Kernel

The kernel of a homomorphism is the set of elements in the domain that are mapped to the zero element (or identity) in the codomain. It measures the failure of a map to be injective and is an essential concept for understanding the structure of a map, particularly in the context of exact sequences where the kernel of one map equals the image of the previous map.

Cokernel

The cokernel of a homomorphism is the quotient of the codomain by the image of the morphism. It measures the failure of surjectivity by 'absorbing' the image, and plays a crucial role in forming new objects in exact sequences. This construction is especially important when studying maps derived from or factoring through cokernels.

Short Exact Sequence

A short exact sequence is a particular type of exact sequence of the form 0 ? A ? B ? C ? 0, where A injects into B and C is the quotient of B by A. This concise framework encapsulates both the injection and surjection properties of the involved maps and is widely used in algebra to study extensions and structure of modules, groups, or other algebraic systems.

Induced Map

An induced map refers to a new map defined on a quotient or subobject that is obtained from an existing map, often via the universal property of quotient objects. In the context of the sequence given in the problem, the maps ? and ? are induced by the original maps, reflecting how the image and kernel relationships transfer to new settings such as cokernels or kernels.

Transcript

00:01 So in the given question we are told if tan of an angle beta is equal to tan of an angle beta is equal to n times sine alpha times cos alpha divided by 1 minus n times cos square alpha then what is the expression for tan of alpha plus beta? so we have an identity to find tan of alpha plus beta which is equal to tan of alpha plus tan of beta tan of alpha plus tan of beta divided by 1 minus tan of alpha times tan of beta.

00:58 So using this identity over here what we can do is we can first simplify the expression for tan beta and substitute it in this equation right so tan beta in the expression on the right hand side of tan beta let's divide the both the numerator and the denominator with cos square alpha and what we would have is n times sine alpha times cos alpha divided by cos square alpha and in the denominator we would have 1 minus n times cos square alpha divided by cos square alpha so 1 divided by cos square alpha minus n times cos square alpha divided by cause square alpha right so this is what we would have and now what we can do is the the cost alpha and cost alpha cancels off in the numerator and we would have n times tan alpha divided by 6 square alpha minus n.

02:07 We have an identity by which we can write that 1 plus tan square alpha, 1 plus tan square alpha is equal to 1 plus tan square alpha is equal to 6 square alpha and using this identity we can substitute for seek square alpha and what we would have is we would have n times tan alpha divided by 1 plus tan square alpha minus 1 right and now now let's substitute this expression of tan beta in the formula of tan of alpha plus beta which we have to find.

03:02 So then what we would have is tan of alpha plus beta is equal to tan of alpha plus n times tan alpha divided by 1 plus tan square alpha minus n.

03:19 1 plus tan square alpha minus n divided by 1 minus tan alpha times n times tan alpha divided by 1 plus tan square alpha and now let's simplify this expression right so what we would have is we would have tan alpha plus tan cube alpha minus n tan alpha plus n tan alpha plus n tan alpha divided by we would have 1 plus tan square alpha minus n minus n times tan square alpha right so this is what we would have after simplifying the numerator and the denominator 1 plus tans square alpha minus n minus n times tan square alpha...

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