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I have a few questions in mind and I would really appreciate if I can clear some questions bogging my mind. Let $X$ be a topological space.

  1. Is there any loop based at $x_0$ that is not homotopic to a constant function based at $x_0$?
  2. Is every continuous function homotopic to a constant function? Let $Y$ be a topological space, and $f: Y \rightarrow X$ be continuous. For $x_0 \in X$, does a homotopy between $f$ and $e_{x_0}$ exist?
  3. Let $Y$ be a compact, connected, linearly ordered, and has least upper bound property. Let $f: Y \rightarrow X$ be continuous and let $0 \in Y$ be the minimum of $Y$. Suppose $f(0) = x_0$. For this $f$, can I define "reverse" of $f$? (Munkres has definition of reverse as: $\overline{f}(y) = f(1-y)$, but ''$-$'' is not defined unless $Y = \mathbb{R}$).
  4. Given this reverse $f$, is there a way to construct (or show existence of) a homotopy between $f*\overline{f}$ and $e_{x_0}$?

Answer to question 1. follows naturally if 2. is true. However, 2. might be false, so I am leaving it as a separate question.

For number 2, I was thinking of deforming subspace induced by the function to a point, but such visualization cannot be formally defined in a general topological space.

Thank you very much in advance.

James C
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    I notice that you tagged this question algebraic-topology. But have you learned about the fundamental group of a space, or the homology groups of a space, and how to apply them to answer these kinds of questions? – Lee Mosher Jun 21 '20 at 22:17
  • For 1 and 2, it will depend on the space $X$. For instance, the disk $D = {(x,y) ~|~ x^2 + y^2 \leq q}$ satisfies 1, while the circle $S^1 = {(x,y) ~|~ x^2 + y^2 = 1}$ doesn't. Intuitively, this is because the disk is "filled in" while the circle has a hole in it, which prevents the obvious loop from contracting. – HallaSurvivor Jun 21 '20 at 23:21
  • For 3 and 4, to talk about path concatenation at all you need a way to glue two copies of $Y$ together in a way that is still homeomorphic to $Y$. With two copies of the unit interval, we can identify $1$ from the first with $0$ from the second, but for an arbitrary $Y$ it's not at all clear how to do this – HallaSurvivor Jun 21 '20 at 23:28
  • @LeeMosher I am new to this subject, and I have just read a section on fundamental group of space. However, I am not still familiar with the notion of homology groups of space. Hopefully I will get there soon! – James C Jun 22 '20 at 00:24
  • @HallaSurvivor On $S^1$, say I start at (1,0), go counterclockwise and make one full turn, stopping at (1,0). Then, I go back clockwise and stop at (1,0). Is this path still homotopic to $e_{(1,0)}$? – James C Jun 22 '20 at 00:36
  • Just a fact that I noticed; you say the author considers $Y=\mathbb{R},$ but, $\mathbb{R}$ is not compact, a condition required by the $Y$ in your question. Perhaps you meant an interval $[a,b]\subset\mathbb{R}.$ – PCeltide Jun 22 '20 at 03:27
  • @PratikApshinge Yes, Munkres cosiders $Y = \mathbb{R}$, but I wanted to know if the same result holds in any topological space. I got myself clear thanks to help from everyone. – James C Jun 23 '20 at 14:51

1 Answers1

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Let us answer the second question first, the first will follow through:

We call a map homotopic to the constant map to be nullhomotopic. Suppose every map $f:Y\rightarrow X$ is nullhomotopic; then in particular, for $Y=X,$ $1_X:X\rightarrow X$ is nullhomotopic, where $1_X$ is the identity map on $X$.

We say a space $X$ is contractible if it has the homotopy type of a point, which means that its identity map is homotopic to the constant map.

Suppose $X$ is contractible. Then, $e_{x_0}\simeq 1_X,$ with some homotopy $f_t.$ For any map $f:Y\rightarrow X,$ we have $f=1_X\circ f\simeq e_{x_0}\circ f,$ by the homotopy $f_t\circ f;$ but $e_{x_0}\circ f$ is a constant map from $Y$ to $X.$ Hence $f$ is nullhomotopic.

What this means in fact is that if all maps into a space $X$ are nullhomotopic, then it must be contractible. Some examples of a space that are not contractible are the circle $S^1$ and the set $\mathbb{R}-\{0\}$ (the last example is quite easy to see intuitively) and so such spaces, there exist functions which are null-homotopic.

For question 3 and 4 - following the comment of @HallaSurvivor, even the discrete set $\{y_0\}$ is a set that is compact, connected, linearly ordered, and has least upper bound property, (there may be many more complicated examples) so I don't think you can generally define a inverse. As you see, you can do so for the unit interval $[0,1],$ and so if the $Y$ is homeomorphic to $[0,1]$ then we may easily be able to find a reverse.

Suppose you are given a reverse path $\overline{f}$ for a path $f:[0,1]\rightarrow X.$ Then take the map $H:I\times I\rightarrow X$ as: $$H(t,s)=\left\{\begin{array}{ll} f(s), & t\leq \frac{1}{2}-s; \\ \overline{f}(s), & \frac{1}{2}-s\leq t \leq 1-s;\\ e_{x_0}(s), & 1-s\leq t. \\ \end{array}\right.$$ This map is continuous and on close observation is actually a homotopy between $f * \overline{f}$ and $e_{x_0}.$

PCeltide
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