A.6. EXACT SEQUENCES 477

Is f (HCn) = ker(g)? Letting c be a cycle, consider f (c) . By definition of exactness,g( f (c)) = 0. Therefore, [0] = [g( f (c))] = g [ f (c)] showing that f (HCn)⊆ ker(g) .

For the other inclusion let [d]∈ ker(g) . In particular this assumes d is a cycle. I need toshow [d] is in f (HCn). We have [0] = g([d]) = [g(d)] . It follows that g(d) = ∂e for somee∈En+1. Since g is onto, e= g

(d̂)

for some d̂ ∈Dn+1 and so g(d) = ∂e= ∂g(d̂)= g(∂ d̂)

so g(d−∂ d̂

)= 0. Therefore, d−∂ d̂ ∈ ker(g) and so d−∂ d̂ = f (c) for some c∈Cn. Thus

0 = f (∂c) and since f is one to one, it follows ∂c = 0. Thus [d] = f ([c]) ∈ f (HCn). Wejust showed that, as mappings on homology groups, f (HCn) = ker(g) .

So far we have this in terms of homology groups:

HCn

f→ HDng→ HEn , Im( f ) = ker(g)

I want to get this:

HCn+1

f→ HDn+1

g→ HEn+1∆→ HCn

f→ HDng→ HEn · · ·

where ∆ is from Lemma A.6.2.{ker∆ = Img} Let [en+1] ∈ ker∆. I need to show [en+1] is in the image of g. Since

[en+1]∈ ker∆,[

f−1∂g−1en+1]= 0 so f−1∂g−1en+1 = ∂cn+1. Thus ∂g−1en+1 = f (∂cn+1).

Letting g(dn+1) = en+1 using the fact that g is onto, we get ∂dn+1 = f (∂cn+1) . Also∂g(dn+1) = g( f (∂cn+1)) so

g(∂dn+1− f (∂cn+1)) = 0

which implies there exists x for which

f (x) = ∂dn+1− f (∂cn+1) = ∂ (dn+1− f (cn+1))

Therefore, x = f−1∂ (dn+1− f (cn+1)) = ∂ f−1 (dn+1− f (cn+1)) = ∂y and so from theabove,

∂ f (y) = ∂ (dn+1− f (cn+1))

so 0 = ∂

(dn+1−

(∈kerg

f (y)+ f (cn+1)

)). But then dn+1− ( f (y)+ f (cn+1)) is a cycle and

g(dn+1− ( f (y)+ f (cn+1))) = g(dn+1) = en+1

Therefore, [en+1] is indeed in the image of g as was to be shown. Thus ker∆⊆ Img.Now consider [g(dn+1)] for dn+1 a cycle. Is this in ker∆? From the definition of ∆,

[f−1

∂g−1g(dn+1)]=

[f−1

=0∂dn+1

]= 0

since f is one to one. Thus Im(g) = ker∆.Next I need to verify exactness at HCn .{Im∆ = ker f} First let en+1 be a cycle and consider ∆ [en+1]≡

[f−1∂g−1en+1

]. Is it in

ker f ? Is it the case that f(

f−1∂g−1en+1)

is a boundary? This expression is just ∂g−1en+1so this is clearly true. Thus Im∆⊆ ker f .

Next suppose f [cn] = [ f cn] = 0 so [cn] ∈ ker f for cn a cycle. I need to show cn ∈ Im∆.Since [ f (cn)] = 0, it follows that f (cn) = ∂dn+1. By exactness, ∂dn+1 ∈ ker(g) because