1. Jul 19, 2011

### jgens

Consider $h:S^1 \times I \to S^1$ defined by $h(s,t) = s^t$. This is a homotopy from the constant map 1 to the identity function on $S^1$. On the other hand, it's easy to show that degree is homotopy invariant. However, we have $\mathrm{deg}(1) = 0$ while $\mathrm{deg}(id) = 1$. Clearly, I've got something mixed up with how I'm thinking about homotopies. Can anybody shed some light on my error?

2. Jul 19, 2011

### Office_Shredder

Staff Emeritus
The map st is not continuous for t not equal to zero or 1. For example when you're taking t=-1/2, then numbers that look like .9999-.0001i get mapped near -1 and 1 gets mapped to 1.

To visualize this, as t goes from one to zero the image of the map st is smaller and smaller arcs of the circle. Right at the start the first thing you do is break the circle in order to make a non-full arc - this is where the homotopy is broken

3. Jul 19, 2011

### jgens

Aw, I see what I messed up. I figured that $h$ was continuous, but you're right that it's clearly not. Thanks!