The index of a Fredholm map is locally constant

Fix $p\in M$ and charts $\phi$ of $M$ at $p$ and $\psi$ of $N$ at $f(p)$, and write $x_0:=\phi (p)$; the representative $\widehat{f}:=\psi \circ f\circ {\phi }^{-1}$ is $C^1$ near $x_0$ and its derivative is continuous there by [L3].

Conjugation identity: for $x$ near $x_0$, writing $y:={\phi }^{-1}(x)$, the chain rule gives $D\widehat{f}(x)=D\psi (f(y))\circ Df(y)\circ D\phi (y{)}^{-1}$; here $D\psi (f(y))$ and $D\phi (y)$ are bounded linear isomorphisms and depend continuously on $y$ by [L5] and the chain rule applied to $\phi \circ {\phi }^{-1}=\mathrm{i}\mathrm{d}$ and $\psi \circ {\psi }^{-1}=\mathrm{i}\mathrm{d}$.

Since $D\widehat{f}(x_0)$ is Fredholm by [L1], [L2] supplies a real $\delta >0$ such that every bounded operator within distance $\delta$ of $D\widehat{f}(x_0)$ is Fredholm with the same index; by continuity in [step 2.1] and [L3] there is a neighbourhood $V$ of $x_0$ with $\mathrm{\parallel }D\widehat{f}(x)-D\widehat{f}(x_0)\mathrm{\parallel }<\delta$ for $x\in V$.

Hence for every $x\in V$ the operator $D\widehat{f}(x)$ is Fredholm with $\mathrm{ind}D\widehat{f}(x)=\mathrm{ind}D\widehat{f}(x_0)$; translating through the conjugation identity of [step 2.1] and the index invariance recorded in [L1] and [L2] gives that $Df(y)$ is Fredholm with $\mathrm{ind}Df(y)=\mathrm{ind}Df(p)$ for every $y$ in the open neighbourhood ${\phi }^{-1}[V]$ of $p$.

Since $p$ was arbitrary, $p\mapsto \mathrm{ind}Df(p)$ is locally constant; by [L4] it is constant on every connected component of $M$.