# Formalizing the "movie" picture of Homotopy (a potentially equivalent formulation of Homotopy)

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?

• No, you just ned $F$ to be continuous. Continuous maps are not always open. Sep 9, 2020 at 21:41
• 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. Sep 9, 2020 at 21:58
• 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. Sep 9, 2020 at 22:37
• 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. Sep 9, 2020 at 22:40
• 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 $[0,1]$. Sep 9, 2020 at 22:50

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): $$F(p,t) = \begin{cases} q & \quad\text{if 0 \le t < 1} \\ r & \quad\text{if t=1} \end{cases}$$

• 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. Sep 9, 2020 at 22:02
• With that understanding, I updated my question. Sep 9, 2020 at 22:33
• The condition I suggested wasn't quite that $F$ is open. Sep 9, 2020 at 22:37
• Yes, it is the same. See my comment under your question. Sep 9, 2020 at 22:37
• Whoops, even though my condition wasn't equivalent to openess, this is a counterexample, thank you!! Sep 9, 2020 at 22:57