Hatcher chapter 0

Hatcher Chapter 0

1) $f:X\to Y$ induces a bijection between the set of path-components of X and the set of path-components of Y.

Proof. Suppose $X=\bigsqcup_i X_i$ , $Y=\bigsqcup_j Y_j$ , where $X_i, Y_j$ are path-components of $X$ and $Y$ . Let $\hat f:\{ X_i \}\to \{Y_j\}$ be defined as follows: take a point $x_{i0}$ in $X_i$ and let $\hat f(X_i)$ be the $Y_j$ such that $f(x_{i0})\in Y_j$ . Now we show this map is well-defined. Let $x_{i0}, x_{i1}\in X_i$ . Since homotopy equivalences preserve path-connectedness, $f(x_{i0}),f(x_{i1})$ are in the same path-component of Y. Hence $\hat f$ is well-defined. Now we show this map is a bijection. Suppose $x_1,x_2\in X$ are on different path-components of $X$ and $\hat f(x_1)=\hat f(x_2)$ . Then $f(x_1)$ and $f(x_2)$ are path-connected. Since $x_1,x_2$ are on two components, $x_1=gf(x_1))$ and $x_2=gf(x_2)$ are not path-connected, a contradiction to path-connectedness is preserved under homotopy equivalences, so $x_1$ and $x_2$ must be on the same path-component of $X$ . Hence $\hat f$ is injective. Since for every path-component $Y_j$ , $\hat f^{-1}(Y_j)$ is the path-component of $X$ such that $g(y_j)\in X_i$ , where $y_j\in Y_j$ , so $\hat f$ is surjective. Therefore, $\hat f$ is a bijection.

2) $f|_{X_i}$ is a homotopy equivalence.

Proof. Since $f$ is a homotopy equivalence, it has an inverse $g$ such that $gf=id_X$ , $fg=id_Y$ . Then $g|_{f(X_i)}$ is a homotopy inverse of $f|_{X_i}$ since $g|_{f(X_i)}f|_{X_i}=id_{X_i}$ , $f_{X_i}g|_{f(X_i)}=id_{f(X_i)}$ . (every $Y_j=f(X_i)$ for a unique $i$ since $\hat f$ is bijective.)

3) Show (1)(2) when path-component is replaced by component.

Proof. The only differences of the proof would be well-definedness and the injectivity of $\hat f$ , but since connectedness is also preserved by homotopy equivalences, the proof follows by replacing path-connectedness by connectedness.

4) If the components of a space X coincide with its path-components, then the same holds for any space Y homotopy equivalent to X .

Proof. Since the components of a space X coincide with its path-components, by (3), there exists a bijection between the set of path-components of $X$ and the set of components of $Y$ , then by (1), there exists a bijection between the set of path-component of $Y$ and the set of components of $Y$ . Since the components must be path-components, the two sets are equal.

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