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Given a homotopy equivalence $f:X\to Y$ and its homotopy cofibre $C(f)$ (that is cofibre of the cofibration $i_1: X\to Z(f)$, where $Z(f)$ is $X\times I$ glued by $f:X\times \{0\}\to Y$ to $Y$) prove that $C(f)$ it is contractible.

Sorry for just stating the problem, but my efforts are completely fruitless. I thought that it should be analogous to the problem of contractibility of a homotopy fibre of a homotopy equivalence, but my proof doesn't seem to apply here...

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I have been thinking about this and I think that a direct homotopy is not out of reach, but I will present what I believe is a standard argument.

Consider the inclusion $i: X \hookrightarrow Z(f), i(x) = (x, 1)$. We want to show that the cofibre of this map is contractible. Notice that $i$ is a cofibration by construction and a homotopy equivalence by assumption that $f$ was.

Notice $i: X \rightarrow Z(f)$ is a homotopy equivalence $rel(X)$ between $i: X \rightarrow Z(f)$ and $id _{X}: X \rightarrow X$. (I have no idea how to draw the right diagram, but this is just a complicated way of saying that $i \circ id_{X} = i$.)

Since both $i, id_{X}$ are cofibrations, it follows that $i$ is a cofiber homotopy equivalence. (If you are not familiar with this term, see p. 46 of May's "Concise Course in Algebraic Topology). In particular, it has a cofiber homotopy inverse $r: Z(f) \rightarrow X$ and there is a homotopy $H: Z(f) \times I \rightarrow Z(f)$ between $i \circ r$ and $id_{Z(f)}$ that is $rel(X)$ at all times. Thus, we can crush $X \times \{ 1 \} \subseteq Z(F)$ on both sides to obtain a homotopy $\tilde{H}: C(f) \times I \rightarrow C(f)$ between a constant map and $id_{C(f)}$. This proves that $C(f)$ is contractible.

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Thank you very much! It's just what I needed. – savick01 Feb 11 '13 at 17:11

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