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This may be a weird question but I really want to know the answer:

Hatcher p.2:

enter image description here

Why is the name 'cylinder' used in this instance? I don't think that this quotient space, namely the mapping cylinder, is homeomorphic to a cylinder. At the beginning of the chapter, he says that "... it should be read in this informal spirit, skipping bits here and there." So should I just ignore this?

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  • $\begingroup$ Even simpler: have you understood why the mapping cone is a "cone"? :) $\endgroup$
    – Avitus
    Jul 2 '13 at 14:15
  • $\begingroup$ I haven't heard such a thing... $\endgroup$
    – Xena
    Jul 2 '13 at 14:21
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    $\begingroup$ It is another construction in topology which is normally introduced before/with the mapping cylinder. Please refer to en.wikipedia.org/wiki/Mapping_cone_(topology) if you are interested in its definition. $\endgroup$
    – Avitus
    Jul 2 '13 at 14:25
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    $\begingroup$ I'm not sure why this is up for closure. It seems to me that questions asking about the motivation behind notation and nomenclature are perfectly valid on the site. $\endgroup$
    – Dan Rust
    Jul 2 '13 at 15:18
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First, one constructs a cylinder over the space $X$, that is, $X\times I$ (here $I=[0,1]$). Then one maps the top of the cylinder into $Y$. You can think on it as gluing the top $X\times \{1\}$ with the image $f(X)$.

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  • $\begingroup$ Makes sense: so here the space homemorphic to a cylinder is the quotient space obtained from $X\times I$, right? $\endgroup$
    – Xena
    Jul 2 '13 at 14:26
  • $\begingroup$ @Bedi, What is (up to homeomorphism) the mapping cylinder of a constant map $f:S^1\to S^2$? $\endgroup$
    – Sigur
    Jul 2 '13 at 14:42
  • $\begingroup$ It is the quotient space of $(S^1\times I)\cup S^2$ obtained by identifying the top of the cylinder $S^1\times I$ with a point in $S^2$. $\endgroup$
    – Xena
    Jul 2 '13 at 15:40
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    $\begingroup$ In particular, it is homeomorphic to the wedge sum $D\vee S^2$ where $D$ is a closed disk. $\endgroup$
    – Dan Rust
    Jul 2 '13 at 15:52
  • $\begingroup$ Since $D$ is contractible, it has the same homotopy type of $S^2$, which is not homotopy equivalent to the cylinder $S^1\times I$. So they can not be homeomorphic. $\endgroup$
    – Sigur
    Jul 2 '13 at 16:23
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It is called a cylinder only really to help with intuition. It is only homeomorphic to an actual cylinder ($S^1\times I$) when we have a map $f\colon S^1\rightarrow S^1$ from the circle to itself, and $f$ is a homeomorphism.

You will later also come across the notion of a mapping torus which is a mapping cylinder of a map from a space to itself, but where we then identify the 'boundary space' with respect to the map. That is, if $f\colon X\rightarrow X$ is a continuous map, then the mapping torus $\mathcal{M}_f$ is the space $(X\times I)/(x,0)\sim(f(x),1)$.

In the same way, this construction is only homeomorphic to an actual torus if we have a map $f\colon S^1\rightarrow S^1$ from the circle to itself, and $f$ is a homeomorphism.

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