Question about Brouwer degree under uniform convergence. I was wondering the following: Say a smooth sequence $u_k$ on a smooth manifold converges uniformly to the limit $u$. Does $u$ preserve the Brouwer degree of the $u_k$'s?
I also believe this is an elementary (maybe stupid) question, which is why I would also appreciate any recommendations for introductary literature on homotopy/homology/degree theory. 
Thanks in advance!
 A: Hard but broadly applicable answer that completely answers the question: Yes, the space of $C^0$ maps from $M \to N$ is locally contractible, and in particular is locally path-connected. So if $u_k \to u$ converges uniformly, then there is a path going from $u_k$ to $u$ for sufficiently large $k$, and hence $u_k$ is homotopic to $u$. Brouwer degree is a homotopy invariant.
Here is an elementary argument using a possibly different definition of degree than you're used to: $u: M \to N$, a map between oriented closed manifolds, has degree $n$ if for a volume form $\omega$ on $N$, $\int_M u^* \omega = n \int_N \omega$. Suppose $u_k \to u$ in the $C^1$ topology, meaning both that $u_k \to u$ uniformly and so do their derivatives. Then $u_k^*\omega \to u^*\omega$ in the space of $C^1$ differential forms (with the $C^1$ topology). Lastly, integration is continuous on the space of $C^1$ differential forms (actually, even $C^0$ differential forms). This $\int_M u_k^* \omega \to \int_M u^* \omega$, and because all of the $\int_M u_k^*\omega$ must be integer multiples of $\int_N \omega$, they must be identically equal to $\int_M u^*\omega$ for large $k$, as desired.
