If f=(fn): 𝐂 →𝐃 is a chain map, prove, for all n ∈Z, that fn+ Zn(𝐂) ⊆Zn(𝐃) and fn+ Bn(𝐂) ⊆Bn

If f=(fn): 𝐂 →𝐃 is a chain map, prove, for all n ∈Z, that...

Key Concepts

Cycles

In the context of chain complexes, cycles are elements that are sent to zero by the differential map. They form the kernel of the differential at a given degree. The fact that chain maps send cycles to cycles is a consequence of them commuting with the differential, thereby preserving the kernel structure necessary for defining homology classes.

Boundaries

Boundaries are elements in a chain complex that are obtained by applying the differential map to elements of the previous degree. They comprise the image of the differential at that degree. Since chain maps preserve the differential structure, they map boundaries in one chain complex to boundaries in another, allowing for a well-defined induced map on homology.

Chain Complex

A chain complex is a sequence of abelian groups or modules connected by homomorphisms (called differentials) such that the composition of two consecutive maps is zero. This property ensures that the images of the differential maps are contained in the kernels of the next, which is crucial for defining cycles and boundaries in homological algebra.

Chain Map

A chain map between two chain complexes is a collection of homomorphisms that commute with the differentials of the complexes. This means that applying the chain map and then the differential is the same as applying the differential and then the chain map, ensuring that the cycle and boundary structures are preserved, which is essential for inducing homomorphisms on the level of homology.

Transcript

00:01 Okay, f is a function on ab, and we want to show what happens if you subtract the areas that you found with r using the right -hand endpoints minus the area that you found using the left -hand end points.

00:15 So i have to go through the whole thing to find out what r -n is.

00:18 We know that delta x is b minus a over n, and xj is a plus i -deltax.

00:35 The height is f of xj, and so the area is f of xj times b minus a over n.

00:48 And then rn is the sum, i equals 1 to, oops, j equals 1 to n, f of xj, b minus a over n.

01:03 So j equals 1 to n f of a plus i b minus a over n times b minus a oops b minus little a over n and then for l n same delta x same same everything oops that's r n okay now i don't need this i don't need this because they're the same until you get to this part.

01:43 Ln is j equals 0 to n minus 1 f of x j b minus a over n which is j equals 0 to n f of a plus i times b minus a over n okay so rn minus ln equals well they both have b minus a over n in them so let's go ahead and put that out in the front okay and this one i'm going to write as j equals one to n minus one f of a plus i delta x plus the last one which i took off f of a this is at times right right here, okay.

02:55 I took off the nth one, which would have been f of a plus n times b minus a over n, oops, minus.

03:14 Now on the l sub n one, i'm gonna take off the zero one.

03:18 I'm gonna put that at the beginning.

03:20 So it's f of a plus zero times b minus a over n plus the rest of them.

03:30 J equals 1 to n minus 1, f of a plus i delta x all right so here's what happened i factored the b minus a out of in over in out of both of them there it is so then i have this series minus this series but i want them to have the same beginning and ending point so i let this one start at one but i wanted it to end at n minus one so here they are the first n minus one of them and then here's the nth one and then in this series i wanted it to start at one...

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