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If I have two functions $f : X \to Y$ and $g : X \to Y$, they are homotopically equivalent if and only if there exists a continuous function $h : [0,1] \times X \to Y$ so that $h(0, x) = y \leftrightarrow f(x) = y$ and $h(1, x) = y \leftrightarrow g(x) = y$. This is the Wikipedia definition.

However, intuitively, a homotopy is a "path between functions". I think of a path $p$ as just a continuous function $p : [0, 1] \to Z$ in some topological space $Z$.

Is there a nice topology we can impose on $X \to Y$ so that homotopies are just paths in $X \to Y$ (i.e. $[0,1] \to (X \to Y))$.

I asked a question here about topologies we can give to the set of functions from $X \to X$ and the answer I got in the comments was the product topology, but I am not convinced that is a good fit here. IIUC the product topology on $X \to Y$ would ignore the topological structure of $X$ and essentially treat it as an index set.

Greg Nisbet
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    Something like a hypertopology on $X\times Y$ seems like a reasonable choice to me in most cases - for "nice" spaces, the graph $${(a,b)\in X\times Y: f(a)=b}$$ of a continuous $f:X\rightarrow Y$ will be a closed subset of $X\times Y$. But I'm not familiar enough with hypertopologies to be too confident, hence the comment as opposed to an answer. – Noah Schweber Dec 28 '22 at 01:41
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    The correct topology to use is the compact-open topology, though we insist on some topological conditions on our spaces. In a previous answer I wrote here I explained the "currying isomorphism" of the compact open topology (really, a categorical adjunction). Look at property (2.) and take $X$ to be the unit interval $[0, 1]$ (you'll have to shuffle notation around to align with mine). This will yield formalism for homotopies as paths in the function space. – paul blart math cop Dec 28 '22 at 05:13
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    Let me know if you have further questions! Extending this idea takes you to very very interesting places in algebraic topology. For instance, you can look up the adjunction between suspension and loop spaces. – paul blart math cop Dec 28 '22 at 05:15

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