# The Unapologetic Mathematician

## 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.

About these ads

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

No comments yet.