Let $X,Y$ be topological spaces, $\phi:X \times [0,\infty) \to Y$ a continuous map. Denote $\phi_t:X \to Y$ by $\phi_t(x)=\phi(x,t)$ and suppose that $\lim_{t \to \infty} \phi_t:=\phi_{\infty}$ exists. Is it true that $\phi_0,\phi_{\infty}$ are homotopic?
I am mostly interested in the case when $X,Y$ are metric spaces; I think that even in this case, an additional uniform convergence assumption might be needed.
Here is an attempt: Denote $\tan:[0,\frac{\pi}{2}] \to [0,\infty]$ (set $\tan(\frac{\pi}{2})=\infty)$, and define $ \tilde \phi(x,t)=\phi(x,\tan(t))$, where $\phi(x,\infty):=\phi_{\infty}(x)$.
To check $\tilde \phi$ is continuous on the product $X \times [0,\frac{\pi}{2}]$, the only non-trivial case is "at infinity": Suppose $(x_n,t_n) \to (x,\frac{\pi}{2})$. Then $(x_n,\tan(t_n)) \to (x,\infty)$. Hence, if the convergence is "uniform" in a suitable sense, everything should be fine.
So, is it true we really need a "uniform convergence"? Is there an example (over metric spaces) where $\phi_{\infty}$ is not homotopic to $\phi_0$ (when we don't assume uniform convergence)?
It's false.
Let $$X = Y = S^1 = [0, 1] / 0 \sim 1.$$
Define $\phi(x, t) = x^{1 + \tan \frac{\pi}{2}t}$. For $t = 0$, this is just the identity on the unit circle. For $t = 1$, it's a distorted version of the identity, in which the trip around the circle starts out slow, but speeds up at the end.
For every $0 \le x < 1$, we clearly have $$\lim_{t \to \infty} \phi(x, t) = 0$$, because larger and larger powers of a number between 0 and 1 tend to $0$.
What about for $x = 1$? Well, $\phi(1, t) = 1$ for every $t$, so in this case,
$$\lim_{t \to \infty} \phi(1, t) = 1.$$
On the other hand, since $1$ and $0$ are the same in the codomain, this amounts to saying that $$ \phi_\infty (x) = 0 $$ for every $x$.
So in this case, clearly $\phi_0$ and $\phi_\infty$ are not homotopic (one has degree $1$, the other degree $0$).