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A homotopy equivalence induces a bijection between path components

Statement

For a space $X$ , write $\pi_0(X)$ for its set of path components. If $f:X\to Y$ is a homotopy equivalence, then

$f_*:\pi_0(X)\longrightarrow\pi_0(Y),\qquad f_*(P_X(x)):=P_Y(f(x)),$

is a well-defined bijection. A homotopy inverse $g:Y\to X$ induces its inverse $g_*$ .

Facts & Assumptions

Given: A homotopy equivalence $f:X\to Y$ with homotopy inverse $g:Y\to X$ .

[A1]

Path components are the equivalence classes for the relation “joined by a path” (Paths, path-connected spaces and path components).

[A2]

One has $g\circ f\simeq\operatorname{id}_X$ and $f\circ g\simeq\operatorname{id}_Y$ (Homotopy equivalences, homotopy inverses and spaces of the same homotopy type).

[L1]

Product projections are continuous, and a map into a product is continuous exactly when its components are continuous (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice).

[L2]

A map is continuous exactly when preimages of open sets are open (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f(\overline{A}) \subseteq \overline{f(A)}$ ).

[L3]

Precomposition and postcomposition preserve homotopies (Precomposition and postcomposition by continuous maps preserve homotopies, including their relative form).

Proof

technique · direct

1.1

If a path $\gamma:I\to X$ joins $x$ to $x'$ , then $f\circ\gamma$ is continuous because $(f\circ\gamma)^{-1}[V]=\gamma^{-1}[f^{-1}[V]]$ for every open $V\subseteq Y$ ; it joins $f(x)$ to $f(x')$ . Hence points in one path component of $X$ have images in one path component of $Y$ , so $f_*$ is well defined. The same argument defines $g_*$ .

A1L2

1.2

If continuous maps $u,v:X\to Y$ are homotopic, then $u(x)$ and $v(x)$ lie in the same path component for every $x\in X$ : precompose the homotopy by the continuous map from a one-point space selecting $x$ , using [L3]; the resulting homotopy of two maps from a point is exactly a path from $u(x)$ to $v(x)$ .

L1L3A1

2.1

Apply step 1.2 to $g\circ f\simeq\operatorname{id}_X$ . For every $x\in X$ , $g(f(x))$ and $x$ lie in the same path component, so $(g_*\circ f_*)(P_X(x))=P_X(x)$ . Thus $g_*\circ f_*$ is the identity on $\pi_0(X)$ .

step 1.2A2

2.2

Applying step 1.2 to $f\circ g\simeq\operatorname{id}_Y$ similarly gives $f_*\circ g_*=\operatorname{id}_{\pi_0(Y)}$ .

step 1.2A2

3.1

Therefore $f_*$ and $g_*$ are mutually inverse functions, so $f_*$ is a bijection.

step 2.1step 2.2∎

Depends on

Used by

Nothing in the library uses this result yet.

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Sources

A. Hatcher, Algebraic Topology, Section 0 (standard reference, not scraped)