Let $M,M'$ be oriented connected compact smooth manifolds of the same dimension, let $S$ be a smooth manifold, and let $\nu : S\times M\rightarrow M'$ be some smooth map. Let $\nu_s : M\rightarrow M'$ denote $\nu_s(\cdot) = \nu(s,\cdot)$ for $s\in S$.

If I'm not mistaken, we can formalize the claim that the per-$s$ degree map $s \mapsto \mathrm{degree}(M\xrightarrow{\nu_s} M')$, as a map $S \rightarrow \mathbb{Z}$, should depend continuously on $s \in S$ (hence be locally constant) by using e.g. de Rham theory.

(Details: Fix some top-degree differential form $\omega$ on $M'$ which is non-exact; then $\mathrm{deg}(\nu_s) = (\int_M \nu_s^*\omega)/(\int_{M'}\omega)$, which we can probably argue depends continuously on $s$.)

More generally, if $M,M'$ are Poincaré duality spaces (with chosen fundamental classes), and $S$ is a topological space and "smooth" everywhere above is replaced by "continuous", does the same claim hold? How can the proof be formalized?


2 Answers 2


What you are considering is the adjoint map $f:S \rightarrow \operatorname{Map}(M, M')$ and asking if for any $x,y \in S$, $\operatorname{deg}(f(x))=\operatorname{deg}(f(y))$. Since the degrees of homotopic maps are equal, it suffices to find a path from $f(x)$ to $f(y)$, but by assumption $M$ is connected, so it is path connected. So we may take the image of any path $x$ to $y$ under $f$.

The same argument works for Poincare duality spaces.

  • $\begingroup$ Thank you! As a technical point: would we need to assume we are working in a "nice" or "convenient" category of topological spaces so that we can apply the Cartesian closed structure, arising from the equipping the mapping space with the compact-open topology? $\endgroup$ Jul 8, 2022 at 22:38
  • $\begingroup$ If you are taking the proof as written, yes (or at least you just need the correspondence between homotopies and paths). However, you could pretty easily get around it by not actually referring to the mapping space and instead just explicitly writing down the homotopy. $\endgroup$ Jul 8, 2022 at 22:41
  • $\begingroup$ Thank you, I think I understand: we don't actually need to make use of the mapping space here, we just need to show that $\nu_s , \nu_{t}$ are homotopic when $s,t$ are "nearby", to show $\mathrm{deg}(\nu_s)$ is locally constant? $\endgroup$ Jul 8, 2022 at 22:44
  • $\begingroup$ I really wouldn't use the term "nearby", you are just using a path between $s,t$ to construct a homotopy between the functions and then using the fact that homology is a homotopy invariant. $\endgroup$ Jul 8, 2022 at 22:55
  • 1
    $\begingroup$ Right, instead of "nearby" I should have said "in the same path component of S". $\endgroup$ Jul 8, 2022 at 22:56

(I assume that when "continuous" replaces "smooth", $\nu$ should be continuous in $s$ but at least $C^1$ in $m$.)

The first step here is almost certainly to write down the details of your proof that $\deg{\!(\nu_s)}$ is continuous in the smooth case.

Despite trying to learn it with you, I'm not confident in my algebraic topology. I'm going to attempt to fill in the gaps in a possibly error-prone manner.

The denominator is independent of $s$, so it suffices to work with the numerator.
Suppose $s_j\to s$; let $A=\{s_j:j\}\sqcup\{s\}$ and note that $A$ is compact. For any tangent vectors $\{v_k\}_k\in(T_pM)^n$, $$(\nu_{s_j}^*\omega)_p(\{v_k\}_k)=\omega_{\nu_{s_j}(p)}(\{D(\nu_{s_j})v_k\}_k)$$ Since $\nu$ is smooth in both $s$ and $x$, $D(\nu_{s_j})\to D(\nu_s)$ as matrices; likewise $\nu_{s_j}(p)\to\nu_s(p)$. But $\omega$ is smooth: at minimum, it exhibits continuous dependence on basepoint and arguments. Thus $$\nu_{s_j}^*\omega\to\nu_s^*\omega$$ pointwise. In fact, something stronger is true: since $A\times M$ is compact, the convergence is not only pointwise but uniform. Thus $$\int{\nu_{s_j}^*\omega}\to\int{\nu_s^*\omega}$$ (Alternatively, since $A\times M$ is compact, $\sup_j{|\nu_{s_j}^*\omega|}$ is bounded, and we can apply Lebesgue's DCT.)

Now look at that proof again. We don't really use smoothness: we need $\omega$ to depend continuously on basepoint and arguments; we need $D(\nu_s)$ to be continuous in $s$…but we don't need further derivatives. So the same argument still works.

  • $\begingroup$ Thanks Jacob! I will think about this some more. (Also yes, I'm thinking of $\nu$ as varying "continuously in $s$" and "smoothly on $M$".) $\endgroup$ Jul 8, 2022 at 22:39

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