Let $\Delta$ be a simplicial complex on $n$ vertices, and $\phi$ a simplicial map that identifies two vertices $x$ and $y$ of $\Delta$. I want to show that the Betti numbers of $\phi(\Delta)$ cannot increase much from those of $\Delta$.


For instance, trivially $b\_0(\phi(\Delta))\leq b\_0(\Delta)$, and I could prove $b\_1(\phi(\Delta))\leq b\_1(\Delta)+1$. It is simply because $\mathrm{Ker}\_{\phi(\Delta)}\partial\_1\setminus\mathrm{Ker}\_{\Delta}\partial\_1=\phi\circ \partial\_1^{-1}(\{c[x]-c[y]\mid c\in\mathbb{F}\})$. The same argument seems to only yield $b\_k(\phi(\Delta))\leq b\_k(\Delta)+O(n^k)$ for larger $k$.


So my questions are, are results with better dependence on $n$ known for higher-order Betti numbers? Can we bound $b\_2(\phi(\Delta))$ in terms of both $b\_2(\Delta)$ and $b\_1(\Delta)$ (and $n$)? Is it true that the sum of Betti numbers cannot increase much under identifying two vertices?



Change of Betti numbers under simplicial maps
Let $\Delta$ be the complex with vertices
$$
x, y, a, b\_1, b\_2, \dots, b\_k, c\_1, c\_2, \dots, c\_k
$$
generated by the following $2$-simplices:
$$
(a,b\_i,c\_i), (a,b\_i, y), (b\_i, c\_i, y), (a, c\_i, x)
$$
For any individual $i$, these 2-simplices generate a contractible subcomplex $\Delta\_i$ whose realization is $D^2$ because all of the simplices contain $a$. The intersection of the subcomplexes $\Delta\_i$ and $\Delta\_j$ is the pair of edges $(a,x)$ and $(a,y)$. The geometric realization of $\Delta$ is a union of discs, glued along a common edge. It is contractible and $\Delta$ has Betti numbers $1, 0, 0, \dots$


The quotient complex $\phi \Delta$ has vertices $x, a, b\_i, c\_i$ and is generated by $2$-simplices
$$
(a,b\_i,c\_i), (a,b\_i, x), (b\_i, c\_i, x), (a, c\_i, x).
$$
Each value of $i$ gives the boundary of a tetrahedron $T\_i$. The intersection of $T\_i$ with $T\_j$ is the edge $(a,x)$. The geometric realization of $\phi \Delta$ is a union of $k$ 2-spheres, glued along a common edge, and its Betti numbers are $1, 0, k, 0, \dots$.


(This is really $k$ copies of the example you described in a now-deleted comment.)


As a result, the Betti numbers of $\phi \Delta$ aren't bounded by those of $\Delta$.




---


Here is a more general result.


Consider the standard simplex with vertices $(v\_0, \dots, v\_k)$, and let $K$ and $L$ be subcomplexes of it.


Let $X$ be the subcomplex of $(x, v\_0, \dots, v\_k)$ generated by the following simplices:


* $(x)$
* $(v\_0, \dots, v\_k)$
* $(x, c\_1, \dots, c\_l)$ whenever $(c\_1, \dots, c\_l)$ is a simplex of $K$


Then $X$ is the mapping cone of the map $K \to \Delta^k$ and
$$
\widetilde H\_\*(X) \cong \widetilde H\_{\*-1}(K).
$$


Similarly, let $Y$ be the subcomplex of $(y,v\_0,\dots,v\_k)$ whose reduced homology groups are a shift of the homology groups of $L$.


Let $\Delta$ be the union of $X$ and $Y$, as a subcomplex of $(x,y,v\_0,\dots,v\_k)$. The intersection of $X$ and $Y$ is the simplex $(v\_0,\dots,v\_k)$ which is contractible, and so
$$
\widetilde H\_\*(\Delta) \cong \widetilde H\_{\*-1}(K) \oplus \widetilde H\_{\*-1}(L).
$$
The complex $\phi \Delta$ is the subcomplex of $(x, v\_0, \dots, v\_k)$ generated by $(x)$, $(v\_0,\dots,v\_k)$, simplices associated to $K$, and simplices associated to $L$. As a result, it is also a mapping cone: the mapping cone of the map $K \cup L \to \Delta^k$, and so
$$
\widetilde H\_\*(\phi \Delta) \cong \widetilde H\_{\*-1}(K \cup L).
$$
Therefore, it suffices to find complexes $K$ and $L$ with not much homology but whose union has large homology.




---


Now let's do a specific example to show that we can get large growth.


For any $d \geq 0$, let $K$ be the subcomplex generated by the $d$-simplices containing $v\_0$, and let $L$ be the subcomplex generated by the $d$-simplices containing $v\_1$. Then $K$ linearly contracts onto $v\_0$ and $L$ linearly contracts onto $v\_1$, so $\widetilde H\_\*(K) = \widetilde H\_\*(L) = 0$, and so 
$$\widetilde H\_\*(\Delta) = 0.$$
Also, by the Mayer-Vietoris sequence, we have
$$\widetilde H\_\*(\phi \Delta) = \widetilde H\_{\*-1}(K \cup L) = \widetilde H\_{\*-2}(K \cap L).$$


The complex $K \cap L$ is the union of all of the $(d-1)$-simplices of $(v\_2,\dots,v\_k)$--the $(d-1)$-skeleton of $\Delta^{k-2}$. Because of this there is a long exact sequence
$$
0 \to \widetilde H\_{d-1}(K \cap L) \to C\_{d-1}(\Delta^{k-2}) \to \dots \to C\_1(\Delta^{k-2}) \to C\_0(\Delta^{k-2}) \to \Bbb Z \to 0,
$$
the rank of $H\_{d+1}(\phi \Delta)$ is the alternating sum of the ranks, which is the alternating sum of the number of simplices in $\Delta^{k-2}$:


$$
\begin{align\*}
b\_{d+1} &= \text{rank of }H\_{d+1}(\phi \Delta)\\
&= \text{rank of }\widetilde H\_{d-1}(K \cap L)\\
&= num\binom{k-1}{d} - \binom{k-1}{d-1} + \binom{k-1}{d-2} - \dots + (-1)^{d-1} \binom{k-1}{1} + (-1)^d \\
&= \binom{k-2}{d}.
\end{align\*}
$$
In particular, this is polynomial of degree $d$. If $n = k+3$ is the number of vertices and $i = d+1$, then the growth of the $i$'th Betti number can be $O(n^{i-1})$.


