# A question about a statement in the paper of C.T.C. Wall

‎‎Suppose that ‎‎$‎‎X$ ‎is ‎a finite ‎2-dimensional CW-complex with free fundamental group ‎and ‎‎$‎‎‎\phi‎ :K ‎\longrightarrow ‎X‎$ is a map which ‎induces ‎an ‎isomorphism ‎of ‎fundamental ‎groups, where $K$ is a finite bouquet of circles with the wedge point $a$. Consider the mapping cylinder $M=X\bigcup_{\phi} (K \times \{1\})$‎. ‎Denote $\pi_n (M_{‎\phi‎},K \times \{ 1\} )$ by $\pi_n (\phi )$‎. ‎Recall ‎that ‎an ‎element ‎of ‎‎$‎‎\pi_n (‎\phi‎ )$ ‎is ‎represented ‎by a‎ ‎pair ‎of ‎maps ‎‎$‎‎‎\beta ‎:‎\mathbb{S}^{n-1}‎\longrightarrow ‎K‎$‎ ‎and ‎‎$‎‎‎\gamma ‎:‎\mathbb{D}^n ‎\longrightarrow ‎X‎$‎ ‎with ‎‎$‎‎‎\gamma‎|_{‎\mathbb{S}^{n-1}‎}=\phi‎ \circ ‎\beta‎$. In the paper ''Finiteness conditions for CW-complexes'' of C.T.C.Wall in Propositon 3.3, Wall has mentioned that since ‎‎$‎‎\pi_2 (‎\phi‎)$ ‎is a‎ ‎free ‎‎$‎‎‎\mathbb{Z}\pi_1 (X)‎$-module‎, then we can attach 2-cells ‎to ‎‎$‎‎K$, ‎necessarily ‎with ‎trivial ‎attaching ‎maps, to make $\phi$ a homotopy equivalence.

My question is that:
What is Wall's mean concerning ''necessarily ‎with ‎trivial ‎attaching ‎maps''?
If his mean is that 2-cells are wedged to the wedge point a, then how can I get to this fact?

Thank you for your help.

## 1 Answer

The attaching map for a 2-dimensional cell is a map $f: S^1 \to K$, and it determines an element $[f]$ of $\pi_1(K)$ (up to conjugacy). What Wall means by trivial attaching maps is that these elements of $\pi_1(K)$ must be trivial. If they weren't, then the map $\pi_1(K) \to \pi_1(K \cup_f D^2) \to \pi_1(X)$ that attaches the cell would send the element $[f] \in \pi_1(K)$ to the trivial element, which contradicts the fact that the map $\phi$ was an isomorphism on $\pi_1$ in the first place.

The homotopy type of the complex you get by gluing 2-dimensional cells only depends on the homotopy classes of the attaching maps (to prove this you typically use multiple applications of the homotopy extension property). Therefore, up to homotopy equivalence it really is the case that you can just attach 2-dimensional cells to the basepoint. The second homotopy group $\pi_2(K \vee \bigvee S^2)$ is a free $\Bbb Z[\pi_1(K)]$-module generated by the copies of $S^2$, and Wall's construction is sending these to a set of generators for $\pi_2(X)$ over $\Bbb Z[\pi_1(X)]$.

• Your explanation is so perfect. Now I understood Wall's mean. Thank you so much for your answer. Could you please explain the construction of $\pi_n (\phi )$ (generators and relations) as $\mathbb{Z}\pi_1 (X)$-module? – M.Ramana Sep 19 '17 at 15:30
• I mean that assume that we attach 2-cells to $K$, ecessarily ‎with ‎trivial ‎attaching ‎maps, to make $L$ and extend $\phi$ over these cells to get a map $\psi :L\longrightarrow X$ so that $\pi_1 (\psi )=\pi_2 (\psi )=0$ (as Wall has mentioned in his paper, p. 59). Now if $\pi_3 (\psi )$ is free, then how do I describe attaching maps? – M.Ramana Sep 19 '17 at 16:02
• @M.Ramana Every element of $\pi_3(\psi)$ is represented by a pair of a map $\alpha: S^2 \to L$ and a map $\beta: D^3 \to X$ so that $\psi \circ \alpha = \beta|_{S^2}$. You can use the generators of $\pi_3(\psi)$ to get attaching maps $\alpha$ to construct your new space $L \to M$ by attaching 3-cells, and the maps $\beta$ to extend your map from $L$ to a map $M \to X$. – Tyler Lawson Sep 19 '17 at 19:31
• I understood what is the form of elements of $\pi_3 (\psi )$. But I don't know yet the form of generators and relations of $\pi_3 (\psi )$ as $\mathbb{Z}\pi_1 (X)$-module exactly . Could you determine them for me exactly? – M.Ramana Sep 20 '17 at 3:00
• In fact, my question is that if $\pi_3 (\psi )$ is free $\mathbb{Z}\pi_1 (X)$-module, then do 3-cells attach to $L$, necessarily ‎with ‎trivial ‎attaching ‎maps (i.e., are they wedged to L)? For example, you told me in your main answer ''what Wall means by trivial attaching maps is that these elements of $\pi_1 (K)$ must be trivial''. But I couldn't understand relation between trivialness of elements of the form $f:S^1 \longrightarrow K$ and freeness of $\pi_2 (\phi )$. – M.Ramana Sep 20 '17 at 3:18