# Proving that homotopic maps have the same degree

Let $M, N$ be compact, connected, oriented manifolds. The degree of a map $f:M \rightarrow N$ is defined as the integer $k$ which satisfies $\int_{M} f^{*}\omega = k\int_{N}\omega$. Using the fact that homotopic maps, say $f, g$ induce the same map on the De Rham cohomology, we have $\int_{M} f^{*}\omega - \int_{M} g^{*}\omega = \int_{} d\nu$ for $\nu$ some $n-1$ form. Now if the manifold has empty boundary, we can use Stokes's theorem to conclude that this equals 0.

However, what about when the manifold does not have empty boundary? A proof that I found relies on the fact that if you have a homotopy $H(x,t)$ then $t \mapsto H_{t}^{*}\omega$ is continuous, and then using the continuity of the integral, concluding that (since the degree is always an integer), $k(t)=\frac{\int_{M}H_{t}^{*}\omega}{\int_{N}\omega}$is also continuous, hence constant. Is there any other way of proving this?

• Slow down. Degree is ordinarily defined for maps between compact, oriented manifolds without boundary (of the same dimension). With boundary, you need more hypotheses to get something homotopy invariant. (Consider maps from a closed interval to itself.) – Ted Shifrin Aug 19 '14 at 23:41
• @TedShifrin Initially I thought that might be the case, however after checking a few different textbooks, they do not specify that this holds only for manifolds without boundary. In JM Lee's Introduction to Smooth Manifolds we also have a corollary (17.8) that homotopic maps have the same induced cohomology maps, whether or not the manifolds have boundary. So, I'm thinking the above holds, but not quite sure. I do see the problem with the interval though, namely that the degree 1 identity map is homotopic to a constant map... – TheManWhoNeverSleeps Aug 20 '14 at 0:18
• Yes, absolutely, homotopic maps induce the same map on cohomology. But for degree you need to look at relative cohomology or restrict to proper maps or ... something. – Ted Shifrin Aug 20 '14 at 2:01
• What you are missing is that without an appropriate hypothesis, the ratio $\int_M f^* \omega \,\, / \,\, \int_N \omega$ need not be an integer. Therefore, while this ratio does vary continuously under a homotopy of $f$, it need not be constant. – Lee Mosher Aug 20 '14 at 16:21
• The "theorem" you are asking about is simply false for manifolds with boundary. However, it remains true if the map $f$ and its homotopy sends boundary to boundary. This allows one to get the equality of the induced maps $H_n(M, \partial M)\to H_n(N, \partial N)$, where $n$ is the dimension of your manifolds. – Moishe Kohan Aug 27 '14 at 5:43