I hope to be clear in my question, I've been meditating on the definition of Homotopy of two continuous maps and I've come to the following thought:

This is the definition I'm adopting: let $f_0, f_1: X \rightarrow Y$ be two continuous function. We say that $f_0$ is homotopically equivalent to $f_1$ iff there is a continuous function called Homotopy $F: X \times I \rightarrow Y$ such that $F(x,0)=f_0 (x)$ and $F(x,1)=f_1 (x)$, where $I=[0,1] \subset \mathbb{R}$. Intuitively one can think about the homotopy map with a mechanical interpretation by letting $f_0$ be the starting function at the time $t=0$ and $f_1$ be the target function at the time $t=1$ and $f_0$ reaches $f_1$ by mean of continuous deformations given by the map $F(x,t)=f_t (x)$. I was wondering why an homotopy is defined sending a cilinder $X \times I$ into the target domain of both function (in our case $Y$)? how do you visualize (conceptually) the deformation process from $f_0$ to $f_1$ by the action of $F$ on the cylinder $X \times I$?

  • $\begingroup$ You should visualize a family of maps parametrized by $t \in I$. $\endgroup$
    – Frost Boss
    Commented Jun 13, 2016 at 21:12
  • $\begingroup$ Personally, I don't like to think of the interval as representing time, because sometimes I want think of something else in the picture as representing time. For example, sometimes when I think about a homotopy of paths, I find it convenient to think of time as the variable I feed into the paths. E.g a geodesic on a Riemannian manifold in a homotopy class might be the "quickest walk to work", in my head. $\endgroup$
    – Max
    Commented Jun 13, 2016 at 21:13

1 Answer 1


Here's how I visualize a homotopy, broken down into tiny steps of visualization.

I start by thinking just about the domain $X \times [0,1]$. At $X \times 0$, I see one homeomorphic copy of $X$, parameterized by a very explicit homeomorphism $i_0 : X \to X \times 0$ defined by $i_0(x)=(x,0)$. Similarly at $X \times 1$, I see another homeomorphic copy of $X$, parameterized by another explicit homeomorphism $i_1(x)=(x,1)$. As the "time variable" $t$ increases from $0$ to $1$, I see a very nice looking continuous family of copies of $X$, namely $X \times t$, sweeping through $X \times [0,1]$ from $X \times 0$ to $X \times 1$, and parameterized by explicit homeomorphisms that I'll denote $i_t(x)=(x,t)$.

Now I forget about time for a moment, and just think about $X \times [0,1]$ as a single topological space. Perhaps, to aid intuition, I might give it another name, maybe $C = X \times [0,1]$. Next I use whatever intuition I have, in order to imagine or to visualize the continuous function $F : C \to Y$. It might be easiest to consider a simple case like $X = S^1$ and so $C =$the cylinder (the outer curved surface of the solid cylinder), in which case the function $F$ can be visualized as some kind of distorted, weird looking cylinder in $Y$, perhaps crossing itself in bizarre ways.

Now that I've got $F : C \to Y$ in my head, I bring the time variable back in: as $t$ increases from $0$ to $1$, the family of functions $i_t :X \to X \times [0,1]$ can be composed with $F$ to give a family of functions $f_t :X \to Y$ defined by $f_t=F \circ i_t$. Thus, the very nice looking continuous family $i_t : X \to C$ is mapped via $F$ to a distorted, weird looking continuous family $f_t : X \to Y$.

  • $\begingroup$ This is amazing! Thank you $\endgroup$
    – Salvatore
    Commented Jun 14, 2016 at 13:50

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