10
$\begingroup$

The question is motivated by my failed comment to this one.

Let $M$ and $N$ be path connected locally compact, locally contractible metric spaces (you may assume that they are manifolds).

Let $\varphi_{n}:M\to N$ be null-homotopic and convergent to $\varphi:M\to N$ in the compact open topology.

Does it follow that $\varphi$ is null-homotopic?

Note that homotopy between maps is a path in $C(M,N)$ (for nice $M$, $N$), and so what my question asks is whether the path component of a constant map is closed in $C(M,N)$.

I am waaay out of my depth here, but perhaps there is a continuous or positive lower semi-continuous functional on $C(M,N)$ akin to the topological degree such that null-homotopic maps would be the zero-set of that functional?

$\endgroup$
  • $\begingroup$ Maybe try: Take $\phi_0=*$ the constant map and write homotopies $\phi_i \sim \phi_{i+1}$, and run them during time $[1/(i+2), 1/(i+1)]$ this defines a map $M\times (0,1] \to N$ which you can complete(?) to $M\times[0,1] \to N$ placing $\phi$ at time $0$. (I don't anything about topology, so maybe there's some point set horror that makes this not work). $\endgroup$ – Dylan Wilson Apr 3 at 18:44
  • 6
    $\begingroup$ If $M,N$ are manifolds then they have CW homotopy type. Thus if $M$ is compact, then $C(M,N)$ (compact-open topology) has CW homotopy type (this is Milnor). In particular it is locally contractible, so locally path connected, so the path components coincide with the connected components and are clopen. $\endgroup$ – Tyrone Apr 3 at 19:58
  • 2
    $\begingroup$ @Tyrone Local contractibility is not a homotopy invariant. For example, a comb space is homotopy equivalent to a point, but is not locally connected. $\endgroup$ – Connor Malin Apr 3 at 20:29
  • 1
    $\begingroup$ @Tyrone Ah I see, a continuous map induces a continuous map between topologized path components and a homotopy equivalence induces a bijection on path components, so in fact the homotopy equivalence induces a homeomorphism on the topologized path components. $\endgroup$ – Connor Malin Apr 3 at 21:04
  • 1
    $\begingroup$ In this case, as you point out, the local contractiblity statement I made is not necessarily true, but a CW complex admits a (numerable) ambiently contractible open covering (I have in mind that it is a Dold space). Such a covering guarantees that its path components are open. Since such a covering may be pulled back along a homotopy equivalence, this guarantees that anything homotopy equivalent to a CW complex has open path components (although it need not be locally contractible - for instance the comb space). $\endgroup$ – Tyrone Apr 8 at 15:18
5
$\begingroup$

I'll provide a general theorem (then one has to apply it to specific circumstances). There is a micro-dictionary/Notation at the bottom of this note.

B-assumption:   Space $\ N\times N\times[0;1]\ $ is normal.

Every metric space $\ N\ $ satisfies B-assumption.

Notation   Let $\ \mathcal W_N\ $ be the set of all closed neighborhoods of diagonal $\ \Delta_N\ :=\ \{(y\ y):\ y\in N\}\ $ in $\ N\times N.$

Family $\ \mathcal W_N\ $ is a basis of all neighborhoods of the diagonal $\ \Delta_N.$

A-assumption:   Space $\ N\ $ is an ANR, meaning that for every normal space $\ X\ $ and closed subset $\ A\ $ of $\ X,\ $ and for every continuous function $\ f:A\to N\ $ there exists a neighborhood $\ U\ $ of $\ A\ $ and continuous $\ F:U\to N\ $ such that $\ F|A=f.$

Thus, $\ N^2\ $ is an ANR too.

Definition:   Sequence $\ f_n:M\to N\ $ is d-convergent to $\ f:M\to N\ \Leftarrow:\Rightarrow $

$$ \forall_{V\in\mathcal W_N}\exists_{m\in\Bbb N} \forall_{n\ge m}\quad (f_n\triangle f)(M)\, \subseteq V $$

Only continuous functions are meant:

THEOREM   Let sequence $\ f_n:M\to N\ $ be d-convergent to $\ f:M\to N.\ $ Then there exists $\ m\in\Bbb N\ $ such that $\ f_n\ $ and $\ f\ $ are homotopic for every $\ n\ge m.$

PROOF   Diagonal $\ \Delta_N\ $ is an ANR because it is homeomorphic to $\ N.\ $ Also, $\ \Delta_N\ $ is closed in $\ N^2\ $ since $\ N\ $ is Hausdorff.  Thus, there exists $\ U\in\mathcal W_N\ $ and a retraction $\ \rho:U\to\Delta_N\ $ (it is an extension of the identity map on $\ \Delta_N.)$

Consider the function $\ g\ $ from a closed subset of $\ N^2\times[0;1]\ $ into $\ N^2\ $ given as follows:

  • $\ \forall_{y\in N^2}\quad g(y\ 0)\ :=\ y; $
  • $\ \forall_{y\in\Delta_N}\forall_{t\in[0;1]} \quad g(y\ t)\ := y; $
  • $\ \forall_{y\in U}\qquad g(y\ 1)\ :=\ \rho(y). $

The arguments of $\ g\ $ belong to the union of three closed subsets of $\ N\times[0;1],\ $ where the three parts of the definition of $\ g\ $ coincide on the overlaps hence $\ g\ $ is well defined. This $\ g\ $ admits an extension $\ G_0\ $ over a closed neighborhood of its closed $3$-part domain. This neighborhood includes $\ V\times[0;1],\ $ where $\ V\subseteq U\ $ is a closed neighborhood of $\ \Delta_N,\ $ because $\ [0;1]\ $ is compact.

Now, by (very elementary and great) Borsuk's homotopy extension theorem, there is homotopy

$$ H:N^2\times[0;1]\to N^2 $$

such that:

  • $\ \forall_{y\in N^2}\qquad H(y\ 0)\ :=\ y; $
  • $\ \forall_{y\in V}\forall_{t\in[0;1]} \quad H(y\ t)\ := G_0(y\ t); $

Let $\ m\in\Bbb N\ $ and $\ n\ge m\ $ be as in Definition. Let homotopies $\ h_n\ h:M\times[0;1]\to N\ $ be given as

$$ h_n\ :=\ \pi'\circ H\circ ((f_n\triangle f)\times\Bbb I );$$ $$ h\ :=\ \pi''\circ H\circ ((f_n\triangle f)\times\Bbb I );$$

where $\ \pi'\ \pi'':N^2\to N\ $ are the canonical projections, and $\ \Bbb I:[0;1]\to[0;1]\ $ is the identity map. We see that:

$$ \forall_{x\in M}\quad h_n(x\ 0)\ =\ f_n(x); $$ $$ \forall_{x\in M}\quad h(x\ 0)\ =\ f(x); $$ $$ \forall_{x\in M}\quad h_n(x\ 1)\ =\ h(x\ 1). $$

Define $\ \gamma_n:M\to Y\ $ by $\ \gamma_n(x):=h_n(x\ 1)=h(x\ 1).\ $ We see that $\ f_n\ $ is homotopic to $\ \gamma_n\ $ is homotopic to $\ f.\,\ $ Remember (observe) that $\ H\ $ in the expressions for $\ h_n(x\ 1)\ $ and $\ h(x\ 1)\ $ is equal to $\ G_0\ $ (we have $\ (f_n(x)\ f(x))\in V).\ $   End of PROOF


NOTATION

  • For functions $\ f:P\to Q\ $ and $\ g:R\to S,\ $ the cartesian product $\ f\times g:P\times Q\to R\times S\ $ is given by $$ \forall_{(p\ r)\in P\times R}\quad (f\times g)(p\ r)\ :=\ (f(p)\ g(r)\,) $$
  • Let $\ P=R\ $ and $\ \Delta_P:=\{(p\ p): p\in P\}.\ $ Then $\ f\triangle g: P\to Q\times S\ $ is given as follows: $$ f\triangle g\ := (f\times g)\circ \delta_P $$ where $\ \delta_P:P\to P\times P\ $, and $\ \forall_{p\in P}\ \delta_P(p):=(p\ p).$
| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ Could you relate your assumption on $N$ to that of the OP (in particular local contractibility)? Could you state whether you claim you're answering the OP's question? $\endgroup$ – YCor Apr 3 at 23:17
  • 2
    $\begingroup$ Thank you for your answer, but I have a hard time following it. What is $(f\Delta f_n)$? Is it the cartesian product of $f$ and $f_n$? If it is, is the $d$-convergence the same as uniform convergence in the case when $N$ is metric? When you apply Borsuk's theorem, what is $Y$? Is it supposed to be $N$? Also, which homotopy are you extending? I guess, you assume that $\Delta_N$ is a deformation retract in $U$, right? does it follows from being ANR? $\endgroup$ – erz Apr 4 at 4:52
  • 1
    $\begingroup$ @erz, I'll add a tiny dictionary. And I'll fix the typo you've spotted, thank you. (It should be N not Y). $\endgroup$ – Wlod AA Apr 4 at 9:55
  • 1
    $\begingroup$ I think now I understand. Thank you! I corrected $Y$ to be $N$ in 2 more places. However, this solves a slightly different problem from what I asked: the convergence of functions is different, and I cannot see how to reduce my problem to this one. I'll wait a bit, and if nobody comes up with a compact-open version, I'll accept your answer. $\endgroup$ – erz Apr 4 at 21:01
  • 1
    $\begingroup$ @erz, thank you very much for everything, including fixing $\ Y\to N$ translation (the standard notation $\ f:X\to Y,\ $ with $\ Y\in \text{ANR},\ $ got imprinted in my brain, I guess). I hope that my answer provides a good environment for your topic (even if only partial -- it's always partial :) ). $\endgroup$ – Wlod AA Apr 4 at 21:59
2
$\begingroup$

Please see the answer to Annie's question. Non-density of continuous functions to interior in set of all continuous functions

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.