With the following information:
- $X$ is a finite simplicial complex of dimension $n$.
- $W$ is a simplicial subcomplex of $X$ of dimension $d<n$.
- (Edit) All the simplices of dimension less or equal than $d$ are either in $W$ or are a face of another simplex of dimension more than $d$.
- $X$ is homotopy equivalent to a wedge of spheres of dimension $n$, $\vee_\mu\mathbb{S}^n$.
Question: Is is true that $cl(X\setminus W)$ is still homotopy equivalent to $\vee_\mu \mathbb{S}^n$?
I tried to find a subcomplex in $X\setminus W$ that gives the homotopy equivalence, but I cannot finish the argument.
If $X = D^2 \lor S^n$ where $D^2$ is the $2$-disk and with $n> 2$ then $X$ is homotopy equivalent to $S^n$. If $W\cong D^2 $ is embedded into the $D^2$ part of $X$ in such a way that the remainder is a $2$-dimensional annulus, then $cl(X \setminus W) \simeq S^1 \lor S^n$ which shows that the statement you gave is not true in general (without extra assumptions).
Edit: This still works as a counterexample with the (added) new premise by thickening the annulus. In more detail as follows:
Let $A^2$ be an annulus of dimension $2$, then $A^2 \times [0,1]^{n-2}$ is $n$-dimensional. Attach $W: =D^2$ onto the inner part of $A^2\times {(0)}$ and call this complex $D$. By construction everything is the face of a $n$-dimensional sub complex. Now $D \lor S^n \simeq S^n$ and $cl(D\lor S^n \setminus W)\simeq S^n \lor S^1$.