# Equivalent definition of a homotopy of functions

It is well known that given $X,Y$ arbitrarily topological spaces, $I$ the unit interval, and continuous functions $f, g : X \rightarrow Y,$ a homotopy between the functions is a continuous function $H : X \times I \rightarrow Y$, such that $H(x,0) = f(x)$ and $H(x,1) = g(x)$.

It is stated in a lot of textbooks that one can think of such an homotopy as a one-parameter family of functions $\{H_{t}\}$, given by $t \mapsto H_{t}$, where $H_{t}(x) := H(x,t)$.

Given that $I$ is a compact space it is clear that the continuity of the homotopy implies the continuity of the later map. But in general the opposite implication might not follow through, which takes me to my question:

Does anyone have a counterexample of a continuous function $\omega: I \rightarrow M(X,Y)$, where $M(X, Y)$ denotes the space of continuous functions with the compact-open topology, such that the induced map $H: X \times I \rightarrow Y$ given by $\omega (t)(x) := H(x,t)$, is not continuous?

• You are probably aware that such an example can exist only if $X$ is not locally compact. – abx Jul 5 '16 at 18:35
• With regards to this question I recommend "M.H. Escardo and R. Heckmann, Topologies on spaces of continuous functions, Topology Proceedings 26 (2001–2002), no. 2, 545–564." available here. This answer to an MO question also has some useful pointers. – Andrej Bauer Jul 5 '16 at 18:55
• @abx yes, thank you. But I would like to know of a specific example. – Jrnm Jul 6 '16 at 0:39
• @AndrejBauer I'll check it, thanks. – Jrnm Jul 6 '16 at 0:40
• Here is a positive result: If $Z$ is a locally compact space and $X,Y$ are any spaces, then there is a bijection $\hom(X \times Z,Y) \cong \hom(X,M(Z,Y))$. In particular, for $Z=[0,1]$, homotopies may be understood as continuously parametrized paths (even though, as was pointed out, they are not necessarily the same as paths of continuous functions). – HeinrichD Sep 12 '16 at 10:23

The adjoint map $H:X×I\rightarrow Y$ is given by the composite $H:X×I\xrightarrow{1\times\omega}X\times M(X,Y)\xrightarrow{ev}Y$ where $ev$ is the evaluation map. You may as well as look for spaces for which the evaluation map is not continuous, and there are plenty such examples.
• It can happen that the evaluation map is not continuous but the composition with $1 \times \omega$ is, so at the very least one needs to pick an example carefully and then actually check that continuity is broken. – Andrej Bauer Jul 7 '16 at 14:18