"Continuous" dependence of degree of continuous/smooth family of maps 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?
 A: 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.
A: (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.
