Homology change by removing a simplex and its face Let $K$ be an abstract simplicial complex and $\alpha,\beta$ be a $k-1$-dimensional and $k$-dimensional simplex of $K$ respectively, such that:


*

*$\alpha$ is a proper face of $\beta$

*$\alpha$ is not a proper face of any other simplex of $K$ (other than $\beta$).


Let $L=K\setminus \{\alpha,\beta\}$ which is a valid simplicial complex.
Without using the fact that $K$ and $L$ are homotopy equivalent/deformation retract, can we argue (the weak result) that 
$$H_i(K)=H_i(L)$$ for all $i\leq k-2$ or $i\geq k+2$ as follows:
By definition, $H_i=\ker\partial_i/\text{Im}\ \partial_{i+1}$. 
We also have $\partial_i: C_i\to C_{i-1}$, and $\partial_{i+1}: C_{i+1}\to C_{i}$
Since $L$ and $K$ only differs by $\alpha$ and $\beta$ which are in dimensions $k-1$ and $k$, $C_i(K)$ and $C_i(L)$ only differ when $i=k-1$ or $i=k$. Hence, $\partial_i$ for $K$ and $L$ only differs when $k-1\leq i\leq k+1$?
Is the above argument valid?
I am concerned about the kernel  $\ker \partial_{k+1}: C_{k+1}\to C_k$, is this different or same for $K$ and $L$? 
Any chance of improving further (without resorting to using homotopy equivalence)?
Thanks a lot.
 A: The following is the simplest proof I could come up with. One may also argue very explicitly in terms of the kernel and image, but it is ultimately a retranslation of this idea.
$L$ is a simplicial subcomplex of $K$. Therefore, the natural map $C_*(L) \to C_*(K)$ is a chain map. We have a short exact sequence $$0 \to C_*(L) \to C_*(K) \to C \to 0,$$ where $C$ is the quotient. To be precise, if $a \in C_*(K)$ projects to $[a]$ in $C$, the differential $\partial [a]$ is the projection of $\partial a$ to $C$.
Because $L = K \setminus \{\alpha, \beta\}$, the chain complex $C$ is concentrated in degrees $k$ and $k-1$, where $C_k = C_{k-1} = \Bbb Z$. In $C_*(K)$, we know that $\partial \beta = \pm \alpha + c$, where $c$ is some sum of simplices other than $\alpha$. 
In particular, $\partial[\beta] = \pm [\alpha]$ in $C$. Therefore, $H_*(C) = 0$ in all degrees. Because the long exact sequence in homology associated to this chain complex is therefore of the form $$0 \to H_*(L) \to H_*(K) \to 0,$$ we see that the inclusion map is an isomorphism.
