The Unapologetic Mathematician

Mathematics for the interested outsider

Homotopic Maps Induce Identical Maps On Homology

The first and most important implication of the Poincaré lemma is actually the most straightforward.

We know that a map f:M\to N induces a chain map f^*:\Omega^k(N)\to\Omega^k(M), which induces a map f^*:H^k(N)\to H^k(M) on the de Rham cohomology. This is what we mean when we say that de Rham cohomology is functorial.

Now if H:f\to g is a homotopy, then the Poincaré lemma gives us a chain homotopy from f^* to g^* as chain maps, which tells us that the maps they induce on homology are identical. That is, passing to homology “decategorifies” the 2-categorical structure we saw before and makes two maps “the same” if they’re homotopic.

As a great example of this, let’s say that M is a contractible manifold. That is, the identity map i:M\to M and the constant map p:M\to\{p\} for some p\in M are homotopic. These two maps thus induce identical maps on homology. Clearly, by functoriality, H^k(i) is the identity map on H^k(M). Slightly less clearly, H^k(\{p\}) is the trivial map sending everything in H^k(M) to 0\in H^k(M). But this means that the identity map on H^k(M) is the same thing as the zero map, and thus H^k(M) must be trivial for all k.

The upshot is that contractible manifolds have trivial homology. And — as an immediate corollary — we see that any compact, oriented manifold without boundary cannot be contractible, since we know that they have some nontrivial homology!

December 6, 2011 - Posted by | Differential Topology, Topology


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  2. […] So what does this mean? The identity map has degree , while we can calculate that the antipodal map has degree . Since these are different, the two maps must not act identically on homology, and therefore cannot be homotopic. […]

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