Consider a matrix $A(p) \in \mathbb{R}^{N \times N}$ that depends linearly on a set of real nonnegative parameters $p =[p_1,\dots,p_n], p_j \geq 0$. Let $A$ be nonnegative irreducible for any nonnegative $p$.
By the Perron-Frobenius Theorem, for any $p$, $A$ has a real simple eigenvalue $\lambda_p$; the characteristic polynomial of $A$ has $\lambda_p$ as a simple root.
It seems from this other question
that the function $\Lambda: p \mapsto \lambda_p$ is a smooth function ($C^\infty$) with respect to $p$.
If $c$ is a regular value of $\Lambda(p)$, then we can use the Implicit Function Theorem to conclude that its preimage $\Lambda^{-1}(c)$ forms an $n-1$ dimensional manifold. But is this manifold connected?
Let $S=[0,1]^2$ and $A(p)=\pmatrix{2+(x-y)&2\\ 2&2-(x-y)}$ for any $p\in S$. Then $A(p)$ is nonnegative and irreducible. Straightforward calculation shows that $$ \Lambda(p)=\rho(A(p))=2+\sqrt{(x-y)^2+4} $$ and its Jacobian matrix is $\pmatrix{\frac{x-y}{\sqrt{(x-y)^2+4}}&\frac{y-x}{\sqrt{(x-y)^2+4}}}$. So, every $c=\Lambda(x,y)$ in $\Lambda(S)$ with $x\ne y$ is a regular value. However, $\Lambda^{-1}(c)$ is disconnected because it is the union of the two disjoint line segments $\{(x,y)\in S:\ x-y=\sqrt{c^2-4c}\}$ and $\{(x,y)\in S:\ x-y=-\sqrt{c^2+4c}\}$.