There are many questions in math.stackexchange dealing with the closed topologist's sine curve $T$ and the Warsaw circle (aka quasi-circle) $W$ which contains a copy of $T$, e.g. recently quasi-circle has trivial homology group.
Many of these questions deal with maps $f$ with domain $I = [0,1]$, $\Delta^n$ = $n$-simplex or $S^n$ = $n$-sphere and codomain $T$ or $W$ and ask
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why the the images of such maps cannot contain the complete "oscillating part" of $T, W$.
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why these maps are null-homotopic.
Clearly 2. is a consequence of 1.
My question:
Are there more general spaces $X$ containing a copy $T'$ of $T$ and more general spaces $Y$ than those mentioned above such that all maps $f : Y \to X$ have the property that their images do not contain the complete "oscillating part" of $T'$?
Note that if we know that for some space $X$ as above, we automatically also know it for $T$ because each map $f : Y \to T' \approx T$ can be regarded as a map into $X$.
Best Answer
The following is perhaps the overarching general result you are looking for. Given a topological space $X$, let $X'$ be $X$ equipped with the following finer topology: a subset $U\subseteq X'$ is open if for each $x\in U$ there is an $X$-neighborhood $V$ of $x$ such that $U$ contains the entire path-component of $V$ which contains $x$.
Theorem: Let $Y$ be a locally path-connected space and $f:Y\to X$ be continuous. Then $f$ is also continuous as a map $Y\to X'$.
For instance, when $X$ is your $T$ or $W$, then $X'$ has the topology which "unglues" the sine curve from its limiting line $\{0\}\times[-1,1]$, since a neighborhood of a point on $\{0\}\times[-1,1]$ no longer needs to contain any of the nearby points on the sine curve (since they are not locally in the same path-component). The resulting space is easily seen to be contractible in the case $X=W$ and a disjoint union of two contractible components in the case $X=T$. So, the Theorem immediately implies any map from a connected locally path-connected space to $X$ is nullhomotopic, since it factors through the contractible space $X'$. Also, no compact subspace of $X'$ contains the entire sine curve, and so no map from a compact locally path-connected space to $X$ can contain the entire sine curve.
Proof of Theorem: Let $U\subseteq X'$ be open and $y\in Y$ be such that $f(y)\in U$; we wish to find a neighborhood of $y$ which $f$ maps into $U$. By definition, there is some $X$-neighborhood $V$ of $f(y)$ such that $U$ contains the path-component of $V$ which contains $f(y)$. Then $f^{-1}(V)$ is a neighborhood of $y$, so by local path-connectedness we can find a path-connected neighborhood $W$ of $y$ which is contained in $f^{-1}(V)$. Since $W$ is path-connected, $f$ must map all of $W$ into the path-component of $V$ which contains $f(y)$. Thus $f(W)\subseteq U$, and $W$ is our desired neighborhood.
(In fact, $X'$ itself is locally path-connected, and so this Theorem says the category of locally path-connected spaces is a coreflective subcategory of the category of topological spaces, with $X\mapsto X'$ as the coreflector. You could also make the analogous constructions and prove the analogous Theorem with "path-connected" replaced by "connected" everywhere. This gives a slightly stronger conclusion in the case $X=T$ or $X=W$, since the topology on $X'$ will still be the same but now $Y$ only has to be locally connected. Note though that in general, the $X'$ you get in this way may not be locally connected; to get a locally connected space you have to iterate the construction transfinitely (and this transfinite iteration is the coreflector to the subcategory of locally connected spaces).)