Why are the following two definitions for a homotopy equivalent?

Let $f,g: X \to Y$ be maps. Then $f \simeq g$ if and only if there is a map $G:X \to Y^{I}$ such that $G(x)(0) = f(x)$ and $G(x)(1) = g(x),$ for all $x \in X$.


Suppose maps $f,g:X \to Y$ are maps. We say that $f$ is homotopic to $g$, written $f \simeq g$, if there is a continuous function $F: X \times I \to Y$ such that $$F(x,0), \quad F(x,1) = g(x), \quad \text{and} \quad F(*,t) = *.$$

In the first definition there is no mention of a base point as opposed to the second definition so I am wondering how that is possible.


$Y^I$ is the space of continuous functions from $I$ to $Y$. The topology on it is the natural topology, which is the finest topology for which for every continuous function $H : X\times I \to Y$ there exists a continuous function $\hat{H} : X \to Y^I$ such that $H(x,t) = \hat{H}(x)(t)$. We make it into a pointed space, by defining the point $*_{Y^I} = t \mapsto *_Y$. So for any continuous function $F : X\times I \to Y$ we have a continuous function $\hat{F} : X \to Y^I$, but $\hat{F}$ will only be base point preserving only if $F(*_X,t) = *_Y$ as that means $\hat{F}(*_X) = (t \mapsto *_Y) = *_{Y^I}$. No requirement is made for $F(x,t)$ when $x \neq *_X$. If the first definition only required $G$ to be a continuous function then these definitions would not be equivalent.

However, to go the other way we need to be able to make a continuous function $G' : X\times I \to Y$ given a continuous function $G : X \to Y^I$ such that $G'(x,t)=G(x)(t)$. Here the fact that $I$, the unit interval $[0,1]$, is compact Hausdorff means that it is locally compact which means that such a function always exists. This also makes the natural topology on $Y^I$ the compact-open topology. In other words, all this rambling about "natural topologies" and "locally compact", just means that we are assured that the function $H(x) = t \mapsto H'(x,t)$ is always continuous and that we can always find such a continuous function $H'$, namely $H'(x,t) = H(x)(t)$. It's obvious that we can do this as plain functions, but it is not obvious that they will be continuous.

  • $\begingroup$ What if $G$ is not base point preserving? $\endgroup$ – user19405892 Apr 5 '16 at 23:36
  • $\begingroup$ I strongly suspect that the meaning of "map" is intended, in this case, to be "base point preserving continuous function". If $f$ ($g$) is a base point preserving map, then $G$ will be if $0$ ($1$) is chosen as the base point for $I$. If none of $f$, $g$, or $G$ are base point preserving, then these definitions aren't equivalent, and the base points are just randomly floating around doing nothing. $\endgroup$ – Derek Elkins Apr 5 '16 at 23:55
  • $\begingroup$ Yes, you are right. In the beginning of the book we assumed functions to be based, and I think map here means the same as function. But how does that mean that $G$ is base point preserving? $\endgroup$ – user19405892 Apr 5 '16 at 23:59
  • $\begingroup$ You are correct, I did make a mistake in the comment. $G$ isn't necessarily base point preserving under the conditions I gave in the comment. Instead, the condition of the first block quote make $G$ base point preserving by definition. There may be non-base point preserving continuous functions (i.e. non-maps) that otherwise satisfy the conditions. Those would not be homotopies. $\endgroup$ – Derek Elkins Apr 6 '16 at 0:04
  • $\begingroup$ So for $G(x,t)$ such that $x \in I$ we have that $G(x,t) = *$? Is that what they mean by $G: X \to Y^{I}$? $\endgroup$ – user19405892 Apr 6 '16 at 0:06

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