# The inclusion map is not an embedding?

In 4b it states:

Show that the inclusion map is an embedding.

The solution say the question is faulty and proceeds to say

Using the definition of embedding given by Guillemin and Pollack the inclusion may not be an embedding.

In which case the definition from Guillemin and Pollack is:

embedding - "An immersion that is injective and proper."

and:

proper - "if the preimage of every compact set in $Y$ is compact in $X$. Intuitively, a proper map is one that maps points "near infinity" in $X$ to points "near infinity" in $Y$.

My best guess is that the definition of proper can be violated somehow. Mapping points "near infinity" to doesn't necessarily happen. So I guess this would depend on the parameterization of the the manifold. If parameterizations were chosen such that $\phi$ and $\theta$ placed "infinity" in different locations.

For example, consider 4 parameterizations of a circle that cover semicircles. The "standard basis" of such a circumstance includes the top half from $\theta_1 \in (0, \pi)$, left half from $\theta_2 \in (\frac{pi}{2}, \frac{3\pi}{2})$, the bottom half $\theta_3 \in (\pi, 2\pi)$ and the right half from $\theta_4 \in (\frac{-\pi}{2}, \frac{pi}{2})$. Each parameterization has points just outside of $\theta_i$'s reach. So consider these points to be "infinity".

Choose a different parameterization $\phi_i$ that is $\theta_i$ rotated $45$ degrees. The "infinity" points change and as a results the set isn't (intuitively speaking) proper under the inclusion map. So in essence it depends on the parameterization.

I believe this is correct as I don't believe I can take a compact segment of $\theta_i$ where the preimage is not well defined for every parameterization $\phi_i$, and thus the preimage of a compact set is not again compact?

Is this correct? Is my intuition behind this correct?

(Also, note the exam is not from this year it is from 2014)

## 1 Answer

There are two standard examples, both of which appear in Guillemin and Pollack. One is the one-to-one immersion of $\Bbb R$ into the plane whose image is a figure eight. The other is given by the mapping $f\colon \Bbb R \to S^1\times S^1$ given by $f(t)=(e^{it},e^{i\alpha t})$, where $\alpha/\pi$ is irrational. In both cases, the image fails to be a submanifold.

• Thank you for the answer, but I don't understand how the inclusion map is used here. For example, in the second example you give, you seem to use $f$ without any mention of an inclusion map. – Dair Oct 6 '16 at 3:42
• Fair enough. I meant the inclusion of the images of these maps. If the domain of the inclusion is in fact a closed submanifold, then of course the map will be an embedding. Also, be careful with what you wrote above. If an immersion is one-to-one and proper, then it's an embedding. But the converse needn't hold. Consider the inclusion of an open interval into $\Bbb R$. – Ted Shifrin Oct 6 '16 at 5:35