In the definition of contractible space, why we deform continuous images of the space? Why not deforming subsets? I have recently started learning algebraic topology. I have seen that the definition of contractible space is meant to capture the idea that the space can be continuously shrunk to a point in itself. The actual definition of contractible space "Identity map $\mathrm{id}_X: X \to X$ is homotopic to a constant map" does not come to me as a first suggestion: I can't figure out why we need to consider maps, I think it's enough to continuously deform the ranges of these maps and we don't need to continuously deform the maps themselves. For example, what is the problem of some definition of the following sort?:

A space $X$ is contractible if there is a continuous $f: [0,1] \to \mathcal{P}(X)$ with $f(0) = X$ and $f(1)$ is a singleton.

Now above I said "of the following sort" because I know that this has problems: at least the topology on $\mathcal{P}(X)$ is not defined and some suitable topology must exist on $\mathcal{P}(X)$ so that the continuity of my $f$ captures the intuition of shrinking, but unfortunately I don't know such topology to make my proposed definition more defensible (Is there one?). After all by mentioning it, I want to say that I don't see why we need the things that we are shrinking to be continuous images of $X$.
Added: As discussed in the comments, I think this is the best way to make the qustion most concrete is to ask for a topology on $\mathcal{P}(X)$ that makes the proposed definition work. But primarily I'm curious why people (experts) find it more natural to think of functions rather than subsets; I suspect I am missing some intuition here and I'm willing to get that. So a natural topology on $\mathcal{P}(X)$ that would do the trick, is a great settlement; but if that's not achievable, I would like to know about your intuition on the actual definition.
 A: I do not think  that it is possible to find a "nice" topology on $\mathcal P(X)$. I do not have a proof that it is impossible, but here are some arguments. By a contractible we mean contractible in the map sense.

*

*Such a topology should be based on a universal construction which associates to each $X$ a topology on $\mathcal P(X)$. If we are allowed to  take an individual construction for $X$ separately, then we could do it as follows:
If $X$ is not contractible, take the discrete topology. Then there is no path from $X$ to a singleton.
If $X$ is contractible, take the trivial topology. Then any two subsets of $X$ are connected by a path.
It is obvious that this approach does not make much sense.


*The intuition of "set shrinking" is that we are allowed to shrink the set continuously to a singleton. Look at the circle $S^1$. Define $f : I \to \mathcal P(S^1), f(t) = \{ e^{is} \mid -\pi t \le s \le \pi t\}$ . Then $f(0) = \{1\}$ and $f(1) = S^1$. The shrinking from $f(1)$ to $f(0)$ is certainly continuous in any intuitive sense, and we can certainly define a topology making this precise. But $S^1$ does not deserve to be considered as contractible.


*We can generalize this to arbitrary bounded metric spaces $X$. Choose $x \in X$ and define $f(t) = D_{r(t)}(x)$ = closed neigborhood of $x$ with radius $r(t)$, where $r(t) = td$ with $d =$  diameter of $X$. This continuously shrinks $X$ to a singleton. You may argue that if $X$ has isolated points, then the size of $f(t)$ may jump discontinuously at some $t$, but for nice spaces like the spheres $S^n$ this is impossible.
Therefore I think that "set shrinking" is an inadequate idea to cover the concept of contractibility. If a space $X$ is contractible, then all points  $x \in X$ are moved continuously along a path $u_x$ to a fixed point $\xi  \in X$, and these movements are compatible in the sense that close points $x_1, x_2$ have close paths $u_{x_i}$. The idea of set shrinking allows to rip the space (see the circle-example), and this is bad.
