I've encountered the following theorem (while reading some topology script):
$\textbf{Theorem}$: Let $K$ be an abstract simplicial complex and $\{u, v\}$ it's edge. If link$_K \{u, v\}$ is contractible in del$_K \{u, v\}$ then $K \simeq K' \lor \Sigma ($link$_K \{u, v\})$, where $K' = \{ \tau \in K : \{u, v\} \not\subset \tau \}$.
Firstly, here are necessary definitons:
Let $\sigma \in K$, then del$_K \sigma$ := $\{ \tau \in K : \sigma \not\subseteq \tau\}$, link$_K \sigma$ := $\{ \tau \in K : \tau \cap \sigma = \emptyset , \tau \cup \sigma \in K \}$.
$\Sigma X$ is the suspension of $X$.
Also, i assumed that $X$ being contractible in $Y$, for topological spaces $X$ and $Y$, means that there exists continuos function $H: X \times [0,1] \rightarrow Y$ such that $H(x, 0) = x$ and $H(x, 1) = const$ for all $x \in X$.
Let's consider the following example:
Define $K = \{\emptyset, \{a\}, \{b\}, \{u\}, \{v\}, \{u,v\}, \{a, u\}, \{a, v\}, \{a,b\}, \{u, v, a\}\}.$
$X$ := link$_K\{u, v\} = \{\emptyset, \{a\}\}$ and $Y$:= del$_K\{u, v\} = K \setminus \{\{u,v\}, \{u,v,a\}\}.$
$X$ is apparently contractible in $Y$.
$K' = K \setminus \{u,v,a\}$, but $K \simeq K' \lor \Sigma ($link$_K \{u, v\})$ is not true since these spaces dont have the same fundamental group (draw the picture).
Where am i wrong? If i am not wrong, what is supposed to be the correct version of the theorem?
(*) Later on in the script, author casually uses this theorem to calculate homotopy types of some really complex spaces, but he still uses the version given above. (so i am probably wrong)