OK, this appears to me like perhaps a dumb question. I am reading Allen Hatcher's Algebraic Topology. I've seen bits and pieces of further material here and there before, now I'm restarting from the beginning.
OK, visually I can see why, say, the 2-sphere $S^2$ can not deformation retract onto it's equator. Intuitively, we can't do this without tearing a hole (or rather two holes). Even with visualizing the 'cylinder' $S^2$ x I (well as a volume) for the function $F$ to $S^1$,
$F:S^2 \times I \rightarrow S^1$,
$F(x,t) = f_t(x)$ the family of homotopic maps. You can see that this mapping cylinder won't work in trying to identify points $F(x,1)$ to points in the circle.
The mapping cylinder for a map and spaces $f:X \rightarrow Y$, is the quotient space
$[(X \times I) \amalg Y]/((x,1) \sim f(x))$.
Or even $S^1$ to its equator, $S^0$ the union of two separate points: not "deformation-retractable". Again, looking at $S^1 \times I$ doesn't appear to have a cont. function to $\{x\} \cup \{y\}$ for distinct points $x,y$ in $S^1$. It doesn't seem to have any mapping cylinder as well.
Disclaimer: OBVIOUSLY a circle is NOT homotopic to a point. Just in case anyone gets any wrong ideas as to where I'm going with this. It's just a question on my part 🙂 If anyone could guide me to the a better intuition or visualization to correct the error of what I'm seeing.
However, if we just pick one point $x_0$ from $S^1$, then the mapping cylinder looks just like a cone, with $F(x,0) = x$, $F(x_0, t) = x_0$ (which is a line going down the cone from the circle to the bottom apex), and $F(x,1) = x_0$. And it looks like the mapping cylinder shows how to continuously deform the circle to the point, even if the point is an element of the circle.
I am thinking this "homotopic picture" of what merely looks like, the circle def. retracting to a point, is somehow misleading me, in the sense that I am just missing something or looking at it wrong or etc…
Can anyone elucidate?
Best Answer
First of all, let's clarify one thing. A circle does retract onto a point, because a retract of a circle to a point on it is just a constant map $r : S^1 \to \{p\}$. What you're really asking about is the fact that a circle doesn't deformation retract onto a point.
A deformation retract would be a homotopy $F : S^1 \times I \to S^1$ taking the circle to one of its points, so to deformation retract a circle to a point you'd need to retract it to a point on the circle, via a series of maps $F_t$ that map the circle to itself. Your map seems to be shrinking the circle to a point, which doesn't work because you're moving the rest of the circle off of itself into the "empty space inside of it," which isn't allowed.
In other words, you're viewing the circle as being embedded in the plane, like you'd draw a circle on a piece of paper. But, topologically, the points "inside the circle" don't exist -- there's only the circle itself.