Understanding Homotopies and Degree Invariance in Topology

[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?

 

[Office_Shredder]

The map s^t 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 s^t 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

 

[jgens]

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