Chain map
A chain map1
It follows that each
Proof
Let
so π§ β π π ( π΄ , π ) . Then π π ( π§ ) = 0 and thus π π π π ( π§ ) = π π β 1 π π ( π§ ) = 0 . Now let π π ( π§ ) β π π ( π΅ , π ) so π β π΅ π ( π΄ , π ) for some π = π π + 1 ( π ) . Then π β π΄ π + 1 and thus π π ( π ) = π π π π + 1 ( π ) = π π + 1 π π + 1 ( π ) . π π ( π ) β π΅ π ( π΅ , π )
Chain maps form morphisms in Category of chain complexes.
Footnotes
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German Kettenabbildung β©
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2010, Algebraische Topologie, ΒΆ3.1.4, p. 128 β©