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Simplicial Structures in Topology

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IV.3 The Cap Product 169<br />

Therefore<br />

d ∩Cp+q( f )(c)=d({ f (x0,..., f (xp)}){ f (xp),..., f (xp+q)} .<br />

On the other hand, C p ( f )=dCp( f ) and so,<br />

Cq( f )(C p ( f )(d) ∩ c)=d({ f (x0,..., f (xp)}){ f (xp),..., f (xp+q)} .<br />

If there are i, j for which f (xi)= f (x j),thenCp+q( f )(c)=0 and at least one of the<br />

two assertions<br />

C p ( f )(d)=0 , Cq( f )({xp,...,xp+q})=0<br />

is true; therefore, the theorem holds also <strong>in</strong> this case. �<br />

Theorem (IV.3.5) and what we have previously done enable us to state the follow<strong>in</strong>g<br />

result:<br />

(IV.3.6) Theorem. Let |L|,|K|∈P and a cont<strong>in</strong>uous function f : |K|→|L| be given.<br />

The follow<strong>in</strong>g diagram:<br />

H p (|L|;Z) × Hp+q(|K|;Z) H p ( f ;Z) × 1<br />

��<br />

1 × Hp+q( f ;Z)<br />

is commutative.<br />

��<br />

H p (|L|;Z) × Hp+q(|L|;Z)<br />

∩<br />

H p (|K|;Z) × Hp+q(|K|;Z)<br />

∩<br />

��<br />

Hq(|K|;Z)<br />

Hq( f ;Z)<br />

��<br />

��<br />

Hq(|L|;Z)<br />

Proof. We leave it to the reader. �<br />

Exercises<br />

1. Let f : |K|→|L| be a given map, z ∈ Hn(|K|,Z), andu ∈ H q (|L|,Z). Provethat<br />

2. Prove that<br />

Hn−q( f )(u ∩ Hn( f )(z)) = H q ( f )(u) ∩ z .<br />

(∀z ∈ Hn(|K|,Z)) 1 ∩ z = z .

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