#### Problem

Show that an injective immersion need not be an embedding.

#### Solution

Let's say we've got a piece of string. We can bend the string so that it looks like a figure eight, with the two ends straddling the point where the eight crosses over. To do this, the string doesn't cross over itself, nor does it need to be bent in a funny way. The key thing is to set it down in just the right way so that we *can't tell it was ever anything but a figure eight!*

More formally, suppose we have an open interval $M$ that gets mapped to a figure eight $N$ in $\mathbb{R}^2$ by the function $f:M \rightarrow N$. Suppose that $f$ is defined such that as a point $x$ approaches either of the endpoints of $N$, $f(x)$ becomes arbitrarily close to $f(a)$ (but never reaches it). We can easily find an $f$ that's injective and for which the corresponding tangent maps at each point are injective (the cubic B-spline used to generate the diagram, for instance!). Hence, $f$ is an injective immersion.

Now the question is: is $f$ an embedding? Well, if $f$ were an embedding, then $N$ would be given the subspace topology from $\mathbb{R}^2$. Using this topology, let's consider open sets around $f(a)$. Any such set will be the intersection of an open ball in $\mathbb{R}^2$ around $f(a)$ with $N$. Hence, *all* open sets around $f(a)$ look like little Xs. But if we have an open interval in $M$ containing $a$, the function $f$ will map it to something that looks more like a line segment. Hence, $f$ is not an open map, hence it is not a homeomorphism, hence it is not an embedding!

*This problem was a "cooperative" effort between Keenan and Andrew.*