homotopy type not constant during a homotopy

Question

What is a possibly easy example of a topological space $X$ and a homotopy $H:X\times I\to X$, $H(x,0)=x$ for all $x\in X$, such that the homotopy type of the subspace $h_t(X)=H(X,t)$ is not constant as $t$ varies from $0$ to $1$?

Answer 1

Let $X$ be a closed disk in $\mathbb R^2$. Think about the family of maps where you first crush the top half of the disk down, so that disk becomes just the lower semi-disk. Now grab hold of the "vertices" of the semi-disk (i.e. the points where the lower semi-circle meets the horizontal diameter), or rather small neighborhoods of these vertices, and start to pull them upwards around the upper semi-circle, so the disk resembles a crescent, or some kind of amoeba.

Keep pulling, until the two pieces you are pulling meet one another and meld together; now the disk has deformed into an annulus, and so the homotopy type has changed.

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Slightly more formally, but less intuitively: You can find a surjection from a disk onto an(y) annulus contained in the unit disk; choose one, and call this $h_1$ (thought of as a map from the disk to itself, whose image happens to be an annulus). Let $h_0$ be the identity. The two maps $h_0, h_1$ from the disk to itself are homotopic (since the disk is homotopic to a point, and since any two maps from a point to the disk are homotopic, given that the disk is path-connected). Any homotopy $H$ that joins them is the desired example.

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One might at first guess that $H$ induces a homotopy equivalence between $X$ and $h_t(X)$. But in fact here is what happens:

Let $X_t$ denote $h_t(X)$, let $j_t$ denote $h_t$, but thought of as a map $X \to X_t$, and let $i_t: X_t \to X$ be the inclusion (so $h_t = i_t \circ j_t$). Then $H$ induces a homotopy between $id_X$ and $i_t \circ j_t$, so $i_t \circ j_t$ is the identity in the homotopy category. But it needn't be the case that $j_t \circ i_t$ is the identity in the homotopy category, and so $j_t$ needn't be a homotopy equivalence.

The formula $i_t \circ j_t \sim id_X$ implies that $j_t$ is a monomorphism (from $X$ to $X_t$) in the homotopy category, and $i_t$ behaves (in the homotopy category) like a sort of "projection" from $X_t$ back to $X$. So $X_t$ has to be "at least as complicated" in the homotopy category as $X$.

But the above example shows that it can be more complicated.