# Proof of another Hatcher exercise: homotopy equivalence induces bijection (part II)

This is part 2 of the question I asked here. Can you tell me if this proof is correct? I thought it's better to ask this in a new separate question, the old post was getting a bit long. Many thanks for your help!

Claim: $f$ restricts to a homotopy equivalence from each path-component of $X$ to the corresponding path-component of $Y$.

Proof:

Let $A \in \pi_0(X)$. I claim that $f|_A : A \rightarrow f(A)$ is a homotopy equivalence with $g|_{f(A)}$ such that $f|_A \circ g|_{f(A)} \simeq id_{f(A)}$ and $g|_{f(A)} \circ f|_A \simeq id_A$, given homotopies $$h:[0,1] \times X \rightarrow Y , h(0, ) = g \circ f, h(1, ) = id_X$$ $$\tilde{h}:[0,1] \times Y \rightarrow X , h(0, ) = f \circ g, g(1, ) = id_Y$$

(i) $g|_{f(A)} \circ f|_A \simeq id_A$:

Then $h^\prime := h|_A$ is a homotopy from $g|_{f(A)} \circ f|_A$ to $id_A$: $$h^\prime(0, ) = h(0, )|_A = g \circ f|_A = g|_{f(A)} \circ f|_A$$ and $$h^\prime(1, ) = h(1, )|_A = id_A$$

(ii) $f|_A \circ g|_{f(A)} \simeq id_{f(A)}$:

$h^{\prime \prime} := \tilde{h}|_{f(A)}$ is a homotopy from $f|_A \circ g|_{f(A)}$ to $id_{f(A)}$: $$h^{\prime \prime}(0, )= \tilde{h}(0, )|_{f(A)} = f\circ g|_{f(A)}$$

But $f \circ g|_{f(A)} = f|_A \circ g|_{f(A)}$ because $A$ is a maximal path-connected subset of $X$ and $f,g$ are continuous so $f(A)$ is a maximal path-connected subset of $Y$. If there was a point $f(y) \in f(A)$ such that there is no path from $x$ to $y$ in $A$, this would cause a contradiction so $g(f(A)) \subset A$ and therefore $$f(g(f(A))) \subset f(A)$$ or $$f \circ g|_{f(A)} = f|_A \circ g|_{f(A)}$$

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$f(A)$ does not have to be a path component. Consider $X = Y = \mathbb R$ and $f = 0$, $g = id$. – Alexander Thumm Mar 26 '11 at 16:19

Like Alexander said, it is not the case that $f(A)$ is necessarily a whole component.
Claim: If $f: X \rightarrow X$ is homotopic to the identity and $A$ is a path-component of $X$, then $f\vert_A$ is homotopic to $1_A$.
Proof: Let $H: X \times I \rightarrow X$ be the homotopy from $f$ to $1_X$. Since $A \times I \subset X \times I$ is path-connected, $H(A \times I)$ is contained within some path-component of $X$. Since $H(a, 1) \in A$, the whole image must be in $A$ (by maximality of path-components). One of the basic properties of the subspace topology is that if $F: W \rightarrow Y$ is a map with $F(W) \subset U$, then $F: W \rightarrow U$ is continuous. Therefore the composition $A \times I \hookrightarrow X \times I \rightarrow A$ is continuous and provides the desired homotopy.