0
$\begingroup$

In an informal sense, what does a slice of the Mapping Cylinder look like? There seem to be two feasible choices: i) Just $X$, the slices are exactly the space $X \times \{t\}$ for some $t \in [0,1]$.

ii) Let $f:X \rightarrow Y$ and define a homotopy $f_t(x)$ connecting $f_0 = id$ to $f_1 = f$. Then one imagines that a slice of the mapping cylinder at some $t \in [0,1]$ looks like $f_t(X)$, partway through the deformation from $X$ to $Y$.

I tend to prefer the second interpretation as it has an intuitive visualization and makes several properties immediately clear, such as the homotopy extension property. However, I have been told this is incorrect and that the first interpretation is the one that is correct. This feels wrong as it seems that the mapping cylinder does nothing in this case, whereas in the other visualization the mapping cylinder 'is' the deformation from $X$ to $Y$ and so it feels useful.

$\endgroup$
4
$\begingroup$

The issue I see with the second interpretation is that $f_0 = \mathrm{id}$ is a map $X \to X$, whereas $f_1 = f$ is a map $X \to Y$, so I'm not sure how to think of a homotopy between them.

The slices of the mapping cylinder are copies of $X$ ($\cong X \times \{t\}$), except at $t = 1$, where the cylinder on $X$ is glued to $Y$ along $f$. The image of $X \times \{1\}$ under this gluing might not be homeomorphic to $X$, since it will identify points in $X$ that $f$ maps to the same point in $Y$.

$\endgroup$
2
$\begingroup$

You might want to consider the map $f: S^1 \to S^1 : \theta \mapsto -\theta$. The mapping cylinder of $f$ is actually diffeomorphic to a standard cylinder, but the map $f$ and the identity are not homotopic, so your "partway through the homotopy" description can't possibly be correct.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.