Let $X$ and $Y$ be topological spaces. Assume $X$ is locally contractible and has no dense finite subset. Assume $Y$ is path-connected.

Given $n$ pairs of points $(x_i, y_i)$ where $x_i\in X$ and $y_i\in Y$ for $1\leq i\leq n$ and a continuous map $f:X\to Y$ can we find a continuous map $g:X\to Y$ homotopic to $f$ such that $g(x_i)=y_i$?

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    $\begingroup$ Any self-homotopy equivalence of the Hawaiian earring $H$ has to preserve the origin $o$, so $X = Y = H, f = \operatorname{id}, n = 1, x_1 = o, y_1 \neq o$ is a counterexample., $\endgroup$ – Bertram Arnold Aug 17 at 10:29
  • $\begingroup$ Corrected after the comment. $\endgroup$ – user145520 Aug 17 at 10:54

Let $X$ be the real line with a doubled origin and $Y$ be $\Bbb R$, and let $f$ be the projection map that collapses the two origins $0^+$ and $0^-$ to $0$. Then any map $g: X \to Y$ satisfies $g(0^+) = g(0^-)$ because $\Bbb R$ is Hausdorff. Therefore, $f$ is not homotopic to any map that sends these two points to distinct ones.

Your question is closely related to the inclusion $\{x_1,\dots,x_n\} \subset X$ having the homotopy extension property. In particular, if it is the inclusion of a neighborhood deformation retract, then such homotopies exist. In the example above, each point individually has a contractible neighborhood but the two origins together do not have a neighborhood that retracts back onto them.

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