Given two continuous maps $f,g : X \rightarrow Y$, the standard definition of a homotopy between f and g is that of a continuous map $H: X \times [0,1] \rightarrow Y$ such that $H(x,0)=f(x)$ and $H(x,1)=g(x)$ for every $x \in X$. Is it equivalent to describe a homotopy as a path $t \mapsto h_t$ of continuous maps with $h_0=f$ and $h_1=g$ which is continuous in some topology on $C(X,Y)$?
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2Yes. The compact-open topology is one such topology, I believe. Maybe some expert can say more about exactly which topologies give this result. – Rob Arthan Jun 03 '21 at 22:08
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@RobArthan I see, thank you Rob! – mits314 Jun 03 '21 at 22:27
1 Answers
Thm. If $X$ is locally compact and Hausdorff, the map $$ \Psi\colon C(X\times Z,Y)\to C(Z,C(X,Y)) $$ defined as $\Psi(H)(z)(x)=H(x,z)$, is a homeomorphism when the spaces of continuous functions are given the compact-open topology.
Proof: Lecture 12: Function spaces (part 4).
Apply this result to the case $Z=[0,1]$ to get a homeomorphism $$ \Psi\colon C(X\times[0,1],Y)\to C([0,1],C(X,Y)) $$
Note that the hypothesis on $X$ is automatically fulfilled for paths and loops: In the former case $X$ is an interval $I$ (usually $[0,1]$) and in the latter $\mathbb S^1$. In other words, $$ C(I\times[0,1],X)\to C([0,1],C(I,X))$$ and $$ C(\mathbb S^1\times[0,1],X)\to C([0,1],C(\mathbb S^1,X)) $$ are homeomorphisms for any topological space $X$.
A second theorem shows that when $K$ is compact and $X$ metric, the compact-open topology of $C(K,X)$ agrees with the topology induced by the distance between functions $$ d(f,g)=\sup\{d(f(\zeta),g(\zeta))\mid\zeta\in K\}. $$ Thus, given that $I$, $[0,1]$ and $\mathbb S^1$ are compact, when $X$ is a metric space, $\Psi$ can bee seen as a homeomorphism of metric spaces.
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I was not aware of such result, it seems like a powerful and versatile theorem. And it applies perfectly to answering my questions! Thanks! – mits314 Jun 27 '21 at 10:51
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1Thanks @noname. I've added a further note, in case you are interested. – Leandro Caniglia Jun 27 '21 at 13:42