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I am working through exercise 6 in Chapter 1 one of Fomenko/Fuchs' "Homotopical Topology" which is stated as:

"Show that the cylinder of any continuous map $f: X \to Y$ is homotopy equivalent to $Y$."

They define the cylinder of a continuous map ($Cyl(f)$) to be the quotient space $(X \times I)\coprod Y/((x, 0) \sim f(x))$, and there are several equivalent definitions of homotopy equivalence.

One being that $X \sim Y$ iff there exists continuous maps $f: X \to Y$ and $g: Y \to X$ such that $f\circ g \sim Id_Y$ and $g \circ f \sim Id_X$.

My thoughts so far: Using geometric intuition this problem makes sense. You attach the base of a cylinder to $Y$ and then collapse the cylinder to its base in $Y$. There is a clearly a continuous injection of $Y \to Cyl(f)$, but my problem comes with defining the appropriate continuous map in the reverse direction. Once I figure out how to define the map I would think that there should be some clear way of defining an explicit homotopy between the maps to the identity.

If the above strategy is not fruitful, then the other option I was considering was using one of the other two equivalent definitions of homotopy in the book which involve constructing a bijection between $\pi(Cyl(f), Z)$ and $\pi(Y, Z)$ such that some diagram commutes using some property of the coproduct and quotient space, but I don't want to employ abstract nonsense unless it is absolutely necessary.

I would like to emphasize that I would like hints as to how to proceed and not the answer to the question.

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Hint: The domain of the quotient map used to define $Cyl(f)$ is a disjoint union of two spaces, $(X \times I) \coprod Y$. So, if you can think of two maps, one from $X \times I$ to $Y$, and the other from $Y$ to $Y$, then putting those two maps together you will get a map from $(X \times I) \coprod Y$ to $Y$. Perhaps you will get lucky and they will induce a map from the quotient space $Cyl(f)$ to $Y$.

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    $\begingroup$ So just the coproduct map induced by the universal property obtained from $f(x, t)=f(x)$ and the identity on $Y$? $\endgroup$ – Brandon Thomas Van Over May 30 '18 at 18:55
  • $\begingroup$ Yup, that's the one. $\endgroup$ – Lee Mosher May 30 '18 at 19:02
  • $\begingroup$ Muchas gracias professor Mosher. $\endgroup$ – Brandon Thomas Van Over May 30 '18 at 19:03

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