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For the problem, I am not given any solution so no idea

Prove that any two continuous maps $f,g; I \to X$ such that

$$f(0)=g(0) \in X$$

are homotopic where $I=[0,1]$ is the unit line.

I just thought, well

$I \in \mathbb{R}$ which is convex, so I can have $h(s,t)=(1-t)f(s)+tg(s)$ as my homotopy.

...No? how should

$$f(0)=g(0)$$

come into play to determine the solution? My reasoning would tell us in fact that any paths are homotopic which shouldn't be the case. I don't understand how I can solve this.

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    $\begingroup$ Completely unclear. I guess they mean homotopy rel $0$. $\endgroup$
    – user98602
    Commented May 9, 2016 at 0:38
  • $\begingroup$ I just wrote the question in the way that is exactly written in the past papers I am looking at. The quetsion can't be any clearer now i hope $\endgroup$
    – Kydo
    Commented May 9, 2016 at 0:39
  • $\begingroup$ I mean to say that the point of their $f(0)=g(0)$ demand is unclear. Your question is fine. $\endgroup$
    – user98602
    Commented May 9, 2016 at 0:42
  • $\begingroup$ So are you saying that my answer isn't entirely wrong?? The demand is redundant and does't matter whether it's there or not? $\endgroup$
    – Kydo
    Commented May 9, 2016 at 0:43
  • $\begingroup$ Is $I$ a closed interval? $\endgroup$
    – AJY
    Commented May 9, 2016 at 0:52

1 Answer 1

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If, for example, $X$ is not path-connected and $f(0)$ is in a different path component from $g(0)$, then $f$ and $g$ can't be homotopic. But if $f(0) = g(0) = x$, then you can "homotope" both $f$ and $g$ to be the constant map $I \to X$ with value $x$, and therefore $f$ and $g$ are homotopic.

As @JackLee points out, your construction of a homotopy does not make sense because in general $X$ there is no notion of multiplying points by numbers or adding points together.

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  • $\begingroup$ I really don't know how to tackle these problems...I mean, I often get told "your homotopy/map doesn't make sense" but I just don't know when it makes sense and when it doesn't. You mention that $X$ doesn't have multiplication of addition, but in cases like this ($S^1,D^2\setminus \{(0,0\}$, there is automatically addition mulitplication available (math.stackexchange.com/questions/1775984/…). Why is it that a circle can have multilcation/division/addition but not for some $X$? The $h$ in the answer looks similar to mine.. $\endgroup$
    – Kydo
    Commented May 9, 2016 at 10:28
  • $\begingroup$ *meant to say multiplication OR addition, not "of" sorry $\endgroup$
    – Kydo
    Commented May 9, 2016 at 10:36
  • $\begingroup$ @Kydo What is the multiplication operation on, for example, a genus-2 surface? Just because some topological spaces have a compatible continuous binary operation does not mean they all do... in fact we give them a special name 'topological groups' or more generally 'H-spaces'. $\endgroup$
    – Dan Rust
    Commented May 10, 2016 at 10:03

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