My simple result about homotopy

[fu^2]

omotopy is my personal traslation of omotopia (since i have only one book, sernesi, that tell about omotopia). Is correct?

So i have found a very nice result: Let $(X,d)$ be a metric space such taht $X\sim {p}$ (i.e. $X$ is homotopic to ${p}$), for some $p\in X$ (another question: how i can say "contraibile"?). Then exists an open subset $U=U(p)\sub X$ s.t. $U$ is path connected.

And a simple corollary: The polish comb doesn't omotopic to ${(0,1)}$.


Open problem: If $X$ is a topological space the sentence is true? because my proof uses necessary a metric...

edit: the grammar...

[fu^2]

I think that my proof is correct if i say that $X$ deformed himsefl to ${p}$.

So the result is not too strong !

[fu^2]

ok, and what do you think about my proof?

[apatriarca]

You are right: it should be $U_p = H(X xx (u,1])$. This isn't necessarily open. So I probably have to change the proof a little bit.

[fu^2]

"apatriarca":Let $H$, $U$ and $[u,1]$ as in the post above. Set $U_p = H^{-1}(X xx (u,1])$ which is open because $H$ is continuous. Consider a point $y \in U_p$ and a point $(x_y, t_y) \in X xx (u,1]$ such that $H(x_y, t_y) = y$. The function $g_y : [0,1] \rightarrow U_p$ defined by $g_y(t) = H(x_y, (1-t)t_y + t)$ is a path between $y$ and $p$. This should prove that $U_p$ is path connected.

@fu^2: What's your proof on metric spaces?

ok, but i don't understand this: $U_p = H^{-1}(X xx (u,1])$, since $H:X xx I\to X$ so $H^{-1}(A)={(x,y)\in X xx I : H(x,t)\in A\subset X}$.


Then, my proof:

i will write $alpha_X(t)=R(x,t)$, where $R$ is the homotopy such that $R(x,0)=1_X(x)=x$ and $R(x,1)=c_p(x)=p$ since $X\sim {p}$ iff exists $R$ homotopy between $1_X$ and a constant function $c_p(x)=p$, for some $p\in X$.
In particular we have that $c_p\sim 1_x rel{p}$, but ${p}$ is a single point, i.e. $R(p,0)=R(p,1)=p$ and so we can suppose, without less of generality, that the length of $\alpha_X(t)=R(x,t)$ - - if we seethe function $alpha_X:[0,1]\to X$ - does to zeros when $x\to p$, since $R$ is continuos (i.e. $R(p,t)=1_X(p)=p=c_p$, so i think that if $X\sim {p}$ isn't necessary suppose that $X$ deform himself to ${p}$... or if i want that all this sentence is true is necessary write that $X$ deform himself to ${p}$?...).

We can define $L_{alpha_X}:X\to RR^+$ as $L_{alpha_X}="sup"_{t_1,t_2\in [0,1]}{d(alpha_X(t_1),alpha_X(t_2))}$.

Then if $x\to p$, we have that $L_{alpha_X}\to 0$ since $alpha_X(t)->alpha_P(t)=p$ and $L_{alpha_p}=0$.

Then $AA \epsilon >0$ exist $U_{delta}(p)$ such that $AA x\in U_{delta}(p)$, $L_{alpha_X}
So we take $epsilon_n=1/n$, then $AAn $ exists $barU_n$ such that $\AA x\in U_{delta_n}(p)$, $alpha_X([0,1])\subset barU_n$, since if $x\in U_{delta_n}(p)$, $L_{alpha_X}
Then $diambarU_n\to 0$, so exists $n_0$ such that $\barU_{n_0}=U\sub X$. By $X\sim {p}$ we have that $alpha_X(1)=p$ and $alpha_X(0)=x$, so $x,p\in barU_{n_0}$ and so $alpha_X(t)$ makes a path between $x$ and $p$. The existence of path between two point is a equivalence relation, so using the transitive property we can say that $U$ is path connected.

[apatriarca]

The proof can be probably simplified noting that taking a real $0 < u < 1$, the set $V_u = H^{-1}(X xx (u,1])$ is an open path connected set. The problem here is to prove that there should be at least one $V_u \ne X$. But if we assume that $V_u = X$ for all $u$, then every path $f_x(t) = H(x, t)$ is disconnected. But $[0,1]$ is connected and $f_x$ is continuous so the assumption $V_u = X$ for all $u$ is absurd and the theorem is proved.
The proof in the previous posts gives a stronger result.

[apatriarca]

Let $H$, $U$ and $[u,1]$ as in the post above. Set $U_p = H^{-1}(X xx (u,1])$ which is open because $H$ is continuous. Consider a point $y \in U_p$ and a point $(x_y, t_y) \in X xx (u,1]$ such that $H(x_y, t_y) = y$. The function $g_y : [0,1] \rightarrow U_p$ defined by $g_y(t) = H(x_y, (1-t)t_y + t)$ is a path between $y$ and $p$. This should prove that $U_p$ is path connected.

@fu^2: What's your proof on metric spaces?

[rubik2]

"fu^2":rubik: No, because i want take $U$ proper subset, i.e. $U!=X$. In fact i wrote $subset$ .

ops

[apatriarca]

Let $H : X xx [0,1] \rightarrow X$ be a continuous map such that $H(x,0) = x$, $H(x,1) = p$ and $H(p,t) = p$ (the homotopy relative to $p$ between the identity $id_X$ and the map $x \mapsto p$). If $U$ is an open neighbourhood of $p$, then $H^{-1}(U)$ is an open set in $X xx [0,1]$. Since the natural projection $\pi : X xx [0,1] \rightarrow [0,1]$ is open, then $V = \pi H^{-1}(U)$ is an open set of $[0,1]$ such that $H(X,t) \in U$ for each $t \in V$. Since $1 \in V$, then there should be an interval $[u, 1] \subseteq V$ (it can be easily proved using the Lebesgue lemma). If $U_p$ is the image of $f(x) = H(x, u)$, then, for each $y \in U_p$, the map $g_y : [0,1] \rightarrow U_p$ given by $g_y(t) = H(\bar x, (1 - t)d + t)$ where $\bar x \in f^{-1}(y)$ is a path between $y$ and $p$. Therefore $U_p$ is path connected.

EDIT: No, it doesn't work because the paths $g_y$ are in $U$ (not $U_p$) and I haven't proved that $U_p$ contains an open set containing $p$.

[fu^2]

rubik: No, because i want take $U$ proper subset, i.e. $U!=X$. In fact i wrote $subset$ .

apatriarca: not locally connected, but exist one (or more) $U=U(p)$ open set in $X$ that they are path connected. Locally connected is too strong!

[apatriarca]

So you want to know if a contractible space is locally path connected?

[rubik2]

my idea was to take $X$ himself as $U$, does it work?

[fu^2]

no, my result is different!

this result is trivial (if you use the function $pi_0$ (or in italian funtore )), my result is different: that exist an open subset $U$ path connected.
This fact is not trivial, since the connection in $X$ is made by the homotopy rel ${p}$ between the identity $1_X$ and the "retrazione" $r:Xto {p}$, so if you take a open subset of $X$ is not trivial that the path of the homotopy $R(x,t)$ is in $U$! and in general is false...(i.e. is not obvious that $U$ is path connected) if you take a casual open set of $X$ with $p\in U$.

Do you understand?

[rubik2]

i think that "connesso per archi" translates "path connected" and "omotopia" translates "homotopy". I know (from lessons) this theorem:

Let $X$ be a topological space. If $X ~ {p}$ then X is path connected.

if you want i can write the proof tomorrow.