For now I have seen homotopy being defined as follows.
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If $\phi,\psi:\Omega\to Z$ are continuous maps, we say that they are homotopic if there exists a continuous $F:\Omega \times [0,1]\to Z$ such that $F(\cdot, 0) = \phi(\cdot)$ and $F(\cdot, 1) = \psi(\cdot)$.
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Two topological spaces $X$ and $Y$ are homotopy equivalent if there exist $f:X\to Y$ and $g:Y\to X$ such that $f\circ g$ is homotopic to $\mathrm{id}_Y$ and $g\circ f$ is homotopic to $\mathrm{id}_X$. In particular, homeomorphic spaces are homotopy equivalent.
This way it seemed to me that homeomorphism is a stronger relation than homotopy, but here comes a problem with my understanding of the latter concept. Let $\Bbb S^1 \subseteq\Bbb R^3$ and let $K\subseteq \Bbb R^3$ be treefoil knot. Am I correct in assuming that $\Bbb S^1$ and $K$ are homeomorphic but there does not exist a homotopy between them in $\Bbb R^3$? I mean that if $\phi:\Bbb S^1\to \Bbb R^3$ and $\psi:\Bbb S^1 \to \Bbb R^3$ are embeddings of $\Bbb S^1$ and $K$, then there does not exist a homotopy $F:\Bbb S^1\times[0,1]\to \Bbb R^3$ that connects them? Then how should the intuition "homeomorphism is stronger than homotopy" be understood?
Best Answer
I think the intuition should say "homeomorphism is stronger than homotopy equivalence".
Perhaps the confusion arises from the following: being homeomorphic to something is a property intrinsic to the space, whereas being homotopic refers to a map $f: X \to Y$, so from that point of view we could say that there is an "ambient" space $Y$ where the deformation takes place. Being homotopy equivalent to something is also an intrinsic property of the space.
I should add that any two maps parametrising the unknot and the trefoil respectively will be homotopic (actually any two maps to $\mathbb{R}^3$ will be) but not isotopic. An isotopy between embeddings $f_0,f_1: X \hookrightarrow Y$ is a map $F: X \times [0,1] \to Y$ such that $F(-,i)=f_i$, $i=0,1$ and the map $F(-,t): X \to Y$ is an embedding for all $t \in [0,1]$. This is because any attempt to unknot the trefoil will lead you to cross two strands. While this is allowed for a homotopy, it is not for isotopy, as one of the $F(-,t)$ will fail to be injective (when the strands cross each other).
To be more precise about why unknotting the trefoil through isotopies is not possible, one has to look at invariants of (say smooth) embeddings $S^1 \hookrightarrow \mathbb{R}^3$, that is knots. A very simple knot invariant that will distinguish the unknot and the trefoil is tricolourability.