Show that two embeddings of $M$ into its product are not homotopic

Question

Assume that $M$ is a compact smooth manifold with positive dimension. We have two ways of embedding $M$ into its product with itself. Way I: $ i_1(m) = (m, m)$ and Way II: $i_2(m) = (a, m)$, where $a \in M$. Show that those two maps cannot be homotopic.

If $i_1$ and $i_2$ are homotopic, then I claim $M$ is null homotopic because if $\pi_1 \circ i_1 = id$ and $\pi_1 \circ i_1$ is a constant map. Then it boils down to showing that $M$ cannot be null homotopic. If $M$ is orientable, then its top cohomology is $\mathbb R$, so it cannot be null homotopoic. What if $M$ is not orientable?

$M$ is certainly a manifold, but why don't you mention this? Is it a topological or a smooth manifold? And $M$ null homotopic means $M$ contractible. — Paul Frost, May 23 '19 at 22:31
Also, there's a typo, it's $\pi_1\circ i_{\bf 2}$ which is a constant map. — Berci, May 23 '19 at 22:34
Maybe he means that if i_1 is homotopic to i_2 then pi_1 of i_1 is homotopic to pi_1 of i_2 which is a constant map. — Noel Lundström, May 30 '19 at 14:37

score 3 · Accepted Answer · answered May 23 '19 at 23:01

3

You can use the following result which shows that the homology mod 2 of a non orientable compact manifold is not trivial.

Homology Groups of Non Orientable Manifold

answered May 23 '19 at 23:01

Tsemo Aristide

89,587

score 2 · Answer 2 · answered May 23 '19 at 22:40

2

You have to show that a compact manifold $M$ cannot be contractible. Each contractible manifold is simply connected, and such manifolds are orientable.

answered May 23 '19 at 22:40

Paul Frost

87,968

Show that two embeddings of $M$ into its product are not homotopic

2 Answers2