Homotopy type of union of two simplicial complexes.

Question

Suppose there exist two simplicial complexes $X$ and $Y$ and we know their homotopy types. If $X \cap Y$ consists of edges and vertices(and no higher dimension faces of them) then how can I compute homotopy type of $X \cup_f Y$? $f$ is an identification map.

Answer 1

I will assume that you are working in the setting of injective adjunction spaces $X\cup_f Y$, where $f: A\subset Y\to X$ is a simplicial embedding. I will then drop the substript $f$ and regard $X, Y$ as subcomplexes in $Z= X\cup_f Y$.

It depends on exactly what you want to compute. For instance, if you want to compute $\pi_1$, you would use Seifert- van Kampen's theorem (but be careful since $X\cap Y$ need not be connected even if $X$ and $Y$). Ronnie Brown wrote a lot about this issue (his personal peeve); references to his books are easy to find on MSE.

Computing higher homotopy groups, is much harder, see here.

I will assume now for the sake of simplicity of notation that $X, Y$ are both connected (but not $X\cap Y$). If you assume that each $\pi_1(X\cap Y, z)$ injects in $\pi_1(X,z), \pi_1(Y,z)$ then you would use the fact that the universal cover of $Z$ breaks as the union of "pieces" $\tilde{X}_i, \tilde{Y}_j$, each of which is isomorphic to the universal covering space of $X, Y$, while intersections of pieces are either empty or contractible. (The index sets for $i, j$ are the cosets $\pi_1(Z)/\pi_1(X)$ and $\pi_1(Z)/\pi_1(Y)$ respectively.)

Hence, the homotopy type of the universal covering space of $Z$ is that of the wedge of the spaces $\tilde{X}_i, \tilde{Y}_j$. But you still have to take into account the structure of these abelian groups as $\pi_1(Z)$-modules. Hopefully this helps.