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?
 A: 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:

*

*$f$ is homotopic to $g$;

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