On the fundamental group of a finite CW complex

Question

So let $X$ be a finite CW complex which is connected.

Q1: Is $\pi_1(X)$ necssarily a finitely presented group?

If the answer is yes, then how does prove it. I've tried to prove it using an induction argument but I'm stuck... So every time one glues a cell then one needs to show that this only throws in finitely many new generators and finitely many new relations... If the argument is too involved then I'd like to have a reference.

For every finitely presented group $G=\langle g_1,\ldots g_n| R_1,\ldots R_m\rangle$ one may construct a connected finite CW complex $X$ having only cells of dimensions $\leq 2$ such that $\pi_1(X)\simeq G$. You have one circle for every generator and one 2-cell for every relation.

Q2: So did topologists try to prove results going in the other direction namely, say that $X$ is a finite connected CW complex with $n_i$ cells of dimension $i$ for $i\leq k$. Suppose that we know nothing about the incidence relations of these cells (except that $X$ is connected) then what can we say about the fundamental group of $X$ (outside the fact that it is finitely presented)?

One might ask similar questions where one imposes some incidence relations on the various cells etc.

Answer 1

Let $X$ be a CW-complex, and write $X_k$ for the $k$-skeleton. The cellular approximation theorem says that any based map $S^1\to X$ is homotopic to a cellular map, and that any two cellular maps that are homotopic are homotopic via a cellular homotopy. This means that the map $\pi_1(X_1)\to\pi_1(X_2)$ is surjective, and the map $\pi_1(X_2)\to\pi_1(X)$ is an isomorphism. Thus, we need only deal with $\pi_1(X_2)$.

Next, $X_1$ is just a connected combinatorial graph (with loops and multiple edges allowed). We can choose a maximal tree $T$ in this graph, and let $A$ be the set of edges not in $T$. For each edge $a\in A$ we have a loop $a'$ given by moving from the basepoint to the start of $a$ within $T$, then traversing $a$, then returning to the basepoint through $T$. It is a standard fact that $\pi_1(X_1)$ is freely generated by $\{a':a\in A\}$.

Next, $X_2$ is obtained from $X_1$ by attaching some $2$-cells. Each attaching map is an unbased map $S^1\to X_1$. After choosing a path from the image of the basepoint to the basepoint in $X$ we get an element of $\pi_1(X_1)$, whose conjugacy class is independent of this choice. An argument with van Kampen's Theorem (most easily done one cell at a time) shows that $\pi_1(X_2)$ is obtained from $\pi_1(X_1)$ by killing off these elements. This gives a finite presentation of $\pi_1(X_2)$.

Answer 2

What Richard Kent said, although it probably deserves to be expanded on.

Suppose for simplicity that $X'$ is obtained from a CW complex $X$ by attaching an $n$-disk $D^n = \{x: |x| \leq 1\}$ along a map $S^{n-1} \to X$, where $n > 2$. To see that the inclusion $X \to X'$ induces an isomorphism on $\pi_1$, use the van Kampen theorem. Namely, cover $X'$ by two open sets where one, $U$ is the image of $\{x: |x| < 2/3\}$ and the other, $V$, is the image of $X \cup \{x: |x| > 1/3\}$, both images being taken along the map to the pushout, $X \cup D^n \to X'$.

Notice that $X$ is a deformation retract of $V$, so $\pi_1$ applied to the inclusion $X \to V$ is an isomorphism. Obviously $\pi_1(U)$ is trivial, and so is $\pi_1(U \cap V)$ since $U \cap V \cong S^{n-1} \times (1/3, 2/3)$ and $\pi_1(S^{n-1})$ is trivial. So when you compute $\pi_1(X')$ as the pushout of the diagram of fundamental groups

$$\pi_1(U) \stackrel{\pi_1(i)}{\leftarrow} \pi_1(U \cap V) \stackrel{\pi_1(j)}{\to} \pi_1(V),$$

according to the conclusion of van Kampen's theorem, the result is isomorphic to $\pi_1(X)$.