0
$\begingroup$

A couple of questions. $X$ is a compact manifold, $Z\subset Y$ is a closed submanifold, then

(i) If $Y$ is contractible (i.e. $id_Y$ is homotopic to a constant map), then any map $f: X\rightarrow Y$ is homotopic to the constant map. May I know why this is true?

$X\xrightarrow{f}Y\xrightarrow{id_Y}Y$ and $id_Y\sim \{y\}$, $\{y\}$ being a constant map.

(ii) No compact manifold is contractible. The reason given is let $Z$ be a single point and let $f:X\rightarrow X$ be an identity map. Then since $X$ is contractible, $f$ is homotopic to a constant map and hence $I_2(f,Z) = 0$. But $\#(f(X)\cap Z) = \#Z = 1$ and hence a contradiction.

I know that if $g:X\rightarrow Y$ is homotopic to a constant map, then $I_2(g,Z) = 0$. So, how does $f\sim \{y\}\implies g\sim \{y\}$?. Thanks!

$\endgroup$
6
  • $\begingroup$ You don't use anything of what you said besides Y is contractible.Try using the homotopy to a constant map to give a homotopy of your function to the constant function. $\endgroup$ Commented Mar 11, 2019 at 5:43
  • $\begingroup$ @Travis, I was trying to show $I_2(f,Z) = 0$ and I've proved that if $f:X\rightarrow Y$ is homotopic to a constant map, then $I_2(f,Z) = 0$. So showing $id_Y\sim \{y\} \implies f\sim\{y\}$ would help me to show $I_2(f,Z) = 0$. $\endgroup$
    – user621937
    Commented Mar 11, 2019 at 5:46
  • $\begingroup$ In general if $F:f\sim g$ is a homotopy then $F\circ (h\times 1)$ is a homotopy $f\circ h\sim g\circ h$ for any map $h$ with suitable codomain. Now observe that any map $f:X\rightarrow Y$ has $f=id_Y\circ f$. $\endgroup$
    – Tyrone
    Commented Mar 11, 2019 at 11:24
  • $\begingroup$ And $D^n$ seems like a perfectly reasonable compact manifold that is contractible. $\endgroup$
    – Tyrone
    Commented Mar 11, 2019 at 11:25
  • $\begingroup$ @Tyrone He may mean compact, connected manifold without boundary, in which case the point is the only such manifold. It can be seen by looking at homology with coefficients in $\mathbb{Z}_2$. $\endgroup$ Commented Mar 11, 2019 at 18:18

2 Answers 2

1
$\begingroup$

For (i) this is actually a consequence of the following general result.

Result: If $X$ and $Y$ are topological spaces and $Y$ is contractible then every continuous map from $X$ to $Y$ is homotopic to a constant map.

For (ii) the result you're trying to prove is incorrect as there are plenty compact manifolds that are contractible. For example any convex subset of $\mathbb{R}^n$ is contractible via a straight line homotopy. It's easy to see then that $[0,1]$ is a compact manifold that is contractible.

$\endgroup$
0
$\begingroup$

For (i), it follows from the result Perturbative mentions and noting that being homotopic is a transitive property. Perturbative is also right in saying that there are certainly compact manifolds that are contractible, [0,1] is clearly an example as given, but if we consider only manifolds without boundaries, then (ii) is very much true (in fact, it's exercise 2.4.6 in Guillemin & Pollack).

As the chapter would have you work out, you would start by proving 2.4.4:

If $f:X \longrightarrow Y$ is homotopic to a constant map, show that $I_{2}(f,Z) = 0$ for all complementary dimensional closed $Z$ in $Y$, except perhaps if $\dim X = 0$.

The idea is that we can always push a point on the manifold into the surrounding ambient space with an arbitrarily small deformation, which will allow its preimage to yield the empty set (zero intersection number). Since this arbitrarily small deformation can be made via a straight line homotopy (manifolds are locally path connected, at least in G&P), the result follows since $I_{2}(f,Z) = I_{2}(g,Z)$ for $f \sim g$.

Use this to prove 2.4.5:

If $Y$ is contractible and $\dim Y > 0$, then $I_{2}(f,Z) = 0$ for every $f:X \longrightarrow Y$, $X$ compact and $Z$ closed, $\dim X + \dim Z = \dim Y$.

(Since all maps into a contractible space are homotopic, in particular, any map is homotopic to the constant map.)

Finally, we approach 2.4.6 by assuming, for a contradiction, the existence of a compact, contractible manifold $Y$ (without boundary) that isn't already the single point space. We then take $Z = \{y_{0}\}$ with $y_{0} \in Y$, and by 2.4.5, we have that $I_{2}(Id_{Y},Z) = 0$, but this is a contradiction since $|Id_{Y}^{-1}(Z)| = |Z| = 1$ by nature of the identity, as you've noted.

$\endgroup$
1
  • $\begingroup$ This answer is very much late, but hopefully it'll shed some light on anyone that stumbles upon this boundaryless confusion. G&P mention very much in passing that boundaries won't be considered in the section--at least for me, it wasn't clear that this would be during the entire section unless otherwise stated and not just during the introductory examples. $\endgroup$ Commented Nov 30, 2020 at 7:56

You must log in to answer this question.