The Unapologetic Mathematician

Mathematics for the interested outsider

Functoriality of Cubic Singular Homology

We want to show that the cubic singular homology we’ve constructed is actually a functor. That is, given a smooth map f:M\to N we want a chain map C_k(f):C_k(M)\to C_k(N), which then will induce a map on homology: H_k(f):H_k(M)\to H_k(N).

The definition couldn’t be simpler. We really only need to define the image of a singular k-cube c in M and extend by linearity. And since c:I^k\to M is a function, we can just compose it with f to get a singular k-cube f\circ c:I^k\to N. What’s the (i,j) face of this singular k-cube? Why it’s

\displaystyle(f\circ c)_{i,j}=(f\circ c)\circ I^k_{i,j}=f\circ(c\circ I^k_{i,j})=f\circ c_{i,j}

and so we find that this map commutes with the boundary operation \partial, making it a chain map.

We should still check functoriality. The identity map clearly gives us the identity chain map. And if f:M_1\to M_2 and g:M_2\to M_3 are two smooth maps, then we can check

\displaystyle\left[C_k(g\circ f)\right](c)=(g\circ f)\circ c=g\circ(f\circ c)=\left[C_k(f)\circ C_k(g)\right](c)

which makes this construction a covariant functor.

August 10, 2011 - Posted by | Differential Topology, Topology

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