Full subcomplexes after attaching a 2-cell to a simplicial complex From the proof of Theorem 7.44 in Rotman's book on algebraic topology:

Let $K$ be a simplicial complex. Let $\alpha$ be a closed edge path in $K$ at $v_0$ and let $K_\alpha$ be obtained by attaching a 2-cell along $\alpha$. Define $L_1$ to be the full subcomplex of $K_\alpha$ with vertices $\text{Vert}(K)\cup\{q_0,\dots,q_{n-1}\}$, and define $L_2$ to be the full subcomplex of $K_\alpha$ with vertices $\{r,v_0,q_0,q_1,\dots,q_{n-1}\}$. Note that $L_1\cap L_2$ is the edge $(v_0,q_0)$ and the loop $\{q_0,\dots,q_{n-q}\}$ and that $L_2$ is isomorphic to the full subcomplex of $D(\alpha)$ with vertices $\{r,q_0,\dots,q_{n-1}\}$.

Here are some details on what $D(\alpha)$ is, and how the attaching map works:

To attach $D(\alpha)$, we just identify $p_i$ with $v_i$. Also, a full subcomplex $L$ of $K$ is just one where, if a simplex in $K$ has all its vertices in $L$ as well, then it is a simplex in $L$ too.
This brings me to my question: First, doesn't $L_1\cap L_2$ also contain the 2-simplex $\{q_{n-1},v_0,q_0\}$? Second, how is $L_2$ isomorphic to $\{r,q_0,\dots,q_{n-1}\}$? It seems like $L_2$ has the two extra edges (1-simplices) $\{v_0,q_0\}$ and $\{v_0,q_{n-1}\}$, along with the 2-simplex $\{q_{n-1},v_0,q_0\}$.
 A: You are right. It is obvious that $L_2$ cannot be (simplicially) isomorphic to the full subcomplex of $D(\alpha)$ with vertices $\{r,q_0,\dots,q_{n-1}\}$ - simply because the number of vertices does not agree. But Rotman does not need that to apply Corollary 7.43.
Let $L'_2 \subset L_2$ be the "simplicial circle" decribed by the closed egde path $\alpha' = (v_0,q_0)(q_0,q_1)...(q_{n-2}, q_{n-1})(q_{n-1},v_0)$. This is a subcomplex of $L_2$ and it is easy to see that the inclusion-induced $\pi(L'_2,v_0) \to \pi(L_2,v_0)$ is an isomorphism. You may for example apply Theorem 7.36. Thus $\pi(L_2,v_0)$ is generated by $[\alpha']$ since $\pi(L'_2,v_0)$ is.
Moreover it not correct that $\pi(L_1 \cap L_2, v_0)$ is generated by $[\alpha]$ because $\alpha$ is not a loop in $L_1 \cap L_2$. What he means is that the image of $\pi(L_1 \cap L_2, v_0)$ in $\pi(L_1,v_0)$ is generated by $[\alpha]$. In fact, use elementary moves to see that $[\alpha] = [\alpha']$ in $\pi(L_1,v_0)$.
Now everything works.
