Homotopy is an equivalence relation

Question

I am trying to prove that for all spaces $X$ and $Y$, homotopy is an equivalence relation on the set of maps from $X$ to $Y$. In my book they say if $f: X \to Y$, then $f \simeq_F f$ where $F(x,t) = f(x)$ for all $x \in X$ and $t \in I$. I am wondering how we know that $F$ is a homotopy between $f$ and $f$ since it must be a continuous function $F: X \times I \to Y$ such that $$F(x,0) = f(x), F(x,1) = g(x), \quad \text{and} \quad F(*,t) = *.$$ How do we prove that $F(x,t)$ is a continuous function? Must we show that the preimage of any open set is open?

Answer 1

Answering strictly your question at the end (not the full proof of "homotopy is an equivalence relation" - still leaving that to you to complete).

Consider the function $p:X \times I \to X$, $p(x, t) = x$. Show that THIS function is continuous (preimage of an open set is that set $\times I$, which is open in the product topology). Then $F = f \circ p$ and the composite of continuous functions is continuous.

This is a very common theme in mathematics - try to decompose "harder" problems into smaller, "we know that..." kind of pieces, and then combine them. The smaller pieces aren't that numerous; it's only the immense number of combinations of such things that makes math seem difficult. (OK, and sometimes you do need to come up with clever tricks...)