EDIT : Thank you Lee Mosher for the helpful response. It seems we need continuity somewhere, maybe the following fixes things? I think what I also wanted was that for all $x\in X$, $t\mapsto F(x,t)$ is a continuous path. Given that, is the statement true?

My mental picture for a homotopy between two continuous functions is a movie of one continuous function turning into the other "continuously".

My guess at a formalization of this mental picture (i.e. my guess at an equivalent formulation of Homotopy between continuous functions which captures this idea) is as follows:

Let $X,Y$ be topological spaces. Let $\cal B_X$ and $\cal B_{[0,1]}$ be bases for $X$ and $[0,1]$ respectively. Let $X\xrightarrow {f,g} Y$ be continuous functions. Then $f$ is homotopic to $g$ if and only if there exists a $F:X\times[0,1]\rightarrow Y$ such that for each open set $U\in\cal B_X$ and $A\in\cal B_{[0,1]}$, $F(U\times A)$ is open in $F(X\times A)$ with the subspace topology.

(Note : This condition isn't the same as $F$ is open. Specifically, consider $X=\mathbb{R}$,$Y=\mathbb{R}^2$, $f:x\mapsto (x,0)$, and $g:x\mapsto (0,0)$ then let $F$ be such that for $t\in[0,1/2]$, $(x,t)\mapsto ((1-2t)x,0)$ and for $t\in(1/2,1]$, $F(x,t)=(0,0)$. Then the condition is satisfied.)

Is this true?

  • $\begingroup$ No, you just ned $F$ to be continuous. Continuous maps are not always open. $\endgroup$
    – Knaus
    Sep 9, 2020 at 21:41
  • $\begingroup$ This is a guess as to an equivalent formulation of Homotopy. (I'll edit that so to be clearer!) The suggested condition isn't the same as open. $\endgroup$ Sep 9, 2020 at 21:58
  • $\begingroup$ The suggested condition is indeed the same as $F$ being open. Every open subset of $X \times [0,1]$ is a union of sets of the form $U \times A$. So if the image of every $U \times A$ is open it follows that the image of every open set is a union of open sets and is therefore open. Converselly, if the image of every open set is open then, each set of the form $U \times A$ being open, it follows that the image of $U \times A$ is open. $\endgroup$
    – Lee Mosher
    Sep 9, 2020 at 22:37
  • $\begingroup$ I think there may have been a misreading, I meant $F(U\times A)$ is open in $F(X\times A)$ but not necessarily in $Y$. Specifically, if we have replaced $Y$ with $Y\times Z$ for any (non-empty) topological space $Z$, it shouldn't affect the condition. $\endgroup$ Sep 9, 2020 at 22:40
  • $\begingroup$ Also note that I am asking one to consider the subspace topology for $F(X\times A)$, so specifically, the notion is asymmetric in $X$ and $[0,1]$. $\endgroup$ Sep 9, 2020 at 22:50

1 Answer 1


That's not the correct definition of homotopy, and its not equivalent to the correct definition.

The collection of subsets of $X \times [0,1]$ given by $$\{U \times A \mid U \in \mathcal B_{\mathcal X}, \, A \in \mathcal B_{[0,1]} \} $$ forms a basis for the product topology on $X \times [0,1]$.

So, the correct definition of a homotopy from $f$ to $g$ is a function $F : X \times [0,1] \to Y$ that is a continuous function with respect to the product topology on $X \times [0,1]$ (and the given topology on $Y$), and that satisfies the two further conditions that $F(x,0)=f(x)$ and $F(x,1)=g(x)$.

And, the correct definition of "$f$ is homotopic to $g$" is that there exists a homotopy from $f$ to $g$ (using the definition of homotopy just given).

So, with that out of the way, your question can be rephrased:

Given continuous $f,g : X \to Y$, are the following equivalent:

  1. $f$ is homotopic to $g$;
  2. there exists an open map $F : X \times [0,1] \to Y$ such that $F(x,0) = f(x)$ and $F(x,1) = g(x)$.

No, they are not equivalent, and here is a counterexample. Let $X = \{p\}$ be a 1 point topological space, and let $Y = \{q,r\}$ be a 2 point discrete topological space. Let $f,g : X \to Y$ be defined by $f(p)=q$ and $g(p)=r$. Then $f$ and $g$ are continuous and they are not homotopic.

However, any function with codomain $Y$ is open, because every subset of $Y$ is open. Therefore the following function satisfies (2): $$F(p,t) = \begin{cases} q & \quad\text{if $0 \le t < 1$} \\ r & \quad\text{if $t=1$} \end{cases} $$

  • $\begingroup$ I meant to suggest a reformulation of the definition of Homotopy (between two continuous functions). I edited the question to reflect that. I am curious if the way I phrased it is also true. $\endgroup$ Sep 9, 2020 at 22:02
  • $\begingroup$ With that understanding, I updated my question. $\endgroup$
    – Lee Mosher
    Sep 9, 2020 at 22:33
  • $\begingroup$ The condition I suggested wasn't quite that $F$ is open. $\endgroup$ Sep 9, 2020 at 22:37
  • $\begingroup$ Yes, it is the same. See my comment under your question. $\endgroup$
    – Lee Mosher
    Sep 9, 2020 at 22:37
  • $\begingroup$ Whoops, even though my condition wasn't equivalent to openess, this is a counterexample, thank you!! $\endgroup$ Sep 9, 2020 at 22:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .