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$\DeclareMathOperator\map{map}$I have the following question:

Let $X$ and $Y$ be topological spaces. Let $\map(X,Y)$ denote the space of non-constant continuous functions from $X$ to $Y$. Suppose moreover that each continuous function from $X$ to $Y$ is homotopic to a fixed continuous function $f \colon X \to Y$. In case $f$ is a homeomorphism, is it true that $\map(X,Y)$ is contractible?

I am grateful if anyone has any counterexamples or what conditions must be imposed on $X$ and $Y$ for the question to be true?

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    $\begingroup$ Sorry, the way you've formulated looks funny: every continuous map should be homotopically equivalent to a continuous map? Sounds like an empty condition to me... $\endgroup$
    – user473423
    Commented Sep 14, 2022 at 17:13
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    $\begingroup$ @Echo : I think $f$ is supposed to be fixed :) $\endgroup$
    – Maxime Ramzi
    Commented Sep 14, 2022 at 20:26
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    $\begingroup$ It might be misleading to remove constant functions from $\mathrm{map}(X,Y)$. Moreover if $Y$ is not totally path-disconnected, and if $X$ is not too small, constant functions are homotopic to non-constant functions, so the artefact of removing constant functions will not be helpful. $\endgroup$
    – YCor
    Commented Sep 15, 2022 at 16:50
  • $\begingroup$ There's a subject called "obstruction theory" that gives you an answer to this question. Perhaps look at Whitehead's book on homotopy theory? $\endgroup$
    – Ryan Budney
    Commented Sep 20, 2022 at 15:00

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Here is a negative answer to one precise formulation of your question. Let $F(X,Y)$ denote the space of maps that are not homotopic to a constant map. Then $F(X,Y)$ can be path connected and contain a homeomorphism without $X$ being contractible.

For example, take $X = B \mathbb Z/2$. Then every map $X \to X$ is either homotopy equivalent to a constant map or to the identity map.

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If there exists a homotopy equivalence $f: X\to Y$ and every other $h$ is homotopic to $f$, then $f$ is homotopic to a constant map, so that $X,Y$ are contractible, and therefore so is $map(X,Y)$.

Except if $Y$ is empty, but then so is $X$, and therefore $map(X,Y)$ is still contractible (for a different reason, though)

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  • $\begingroup$ Thanks for your answer, and you are right. It was my mistake not to eliminate constant functions. I forgot to formulate my hypotheses correctly since I am trying to prove something in higher category theory. (In that context, constant functors are not allowed due to the nature of the functors I am working with). But having a topological idea will help me. $\endgroup$
    – Wilson Forero
    Commented Sep 15, 2022 at 16:31

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