Let $M$ be a smooth manifold of dimension greater than $2$.
Suppose $H_1,H_2,\dots,H_k$ are disjoint embedded submanifolds of $M$. Suppose that $H_k$ is open and dense in $M$, and that $\cup_{i=1}^k H_i$ is also an open submanifold of $M$. Let $f:\cup_{i=1}^k H_i \to N$ be a smooth injective map. ($N$ is another smooth manifold).
Finally, assume $f|_{H_i}$ is an immersion for every $i$. Is it true that $f:\cup_{i=1}^k H_i \to N$ is an immersion?
The dimensions of the $H_i$ are distinct, and only the last one, $H_k$ is open. (All the rest have positive codimension in $M$).
Edit:
As Ted Shifrin showed, the answer is negative. The idea is actually quite simple: Even if $H_k$ is open and dense in $M$, the rank of $f$ "can fall in the limit"- that is it can be non-maximal outside $H_k$. The reason for that is that outside $H_k$ we only have "partial injectivity", that is $df$ is is assumed to be injective only on a strict subspace of the tangent space of $M$.
Here is why I thought there should be a counter-example:
Let $p \in \cup_{i=1}^k H_i $. Then there is exactly one $j$, $1 \le j \le k$ such that $p \in H_j$.
Note that $T_p(\cup_{i=1}^k H_i)=T_pM$, since $\cup_{i=1}^k H_i$ is open, by assumption. We ask whether or not $df_p:T_pM \to T_{f(p)}N $ must be injective.
Of course, we know $df_p|_{T_pH_j}$ is injective. Since $p \notin H_i$ for $i \neq j $, I don't see a way to use the assumption $f$ is an immersion on the other $H_i$.
Now that I have the question sorted out, I believe the answer is no.
Take the surface $S=\{y^2=x^3\}$ in $\Bbb R^3$. The $z$-axis immerses just fine, as does the top stratum $H = S - \{z\text{-axis}\}$. But the surface does not immerse. (You need normal data as you approach the lower strata.)
EDIT: To be more specific, let $H_2 = \{(s,t)\in\Bbb R^2: s\ne 0\}$ and $H_1 = \{(0,t)\in\Bbb R^2\}$. Then $H_1\cup H_2 = \Bbb R^2 = M$. Now let $f\colon\Bbb R^2\to\Bbb R^3$ be given by $f(s,t) = (s^2,s^3,t)$. $f$ is an immersion on $H_i$ individually, but fails to be an immersion.