CW complexes

Recall the general definition: a finite two-dimensional CW complex structure for $X$ expresses $X$ as being built in stages $X^0 \subset X^1 \subset X^2 = X$. Here $X^0$ is a finite discrete set of points called $0$-cells. $X^1$ is formed from $X^0$ by attaching finitely many $1$-cells (copies of a $1$-disk) via continuous maps from their boundary into $X^0$. Finally $X = X^2$ is formed from $X^1$ by attaching finitely many $2$-cells via continuous maps from their boundary into $X^1$.

The $2$-sphere $S^2$ has the structure of a CW complex with $X^0$ a single point $e^0$ and $X^1 = X^0$ (there are no $1$-cells). Then $X^2 = X^0 \cup e^2$ is formed by using the constant map sending $\partial e^2$ to the point $X^0$. Verify that $S^2$ is homeomorphic to this CW complex $e^0 \cup e^2$.

For the torus $T$, think of it as formed from a square $D^1 \times D^1$ by identifying $(x,1) \sim (x,0)$, $(1,y) \sim (0,y)$. The four corners are identified to a single $0$-cell $e^0$. The edge $D^1 \times \left\{ 0\right\}$ (identified with $D^1 \times \left\{ 1\right\}$) forms a $1$-cell $e^1_1$, and the edge $\left\{ 1\right\} \times D^1$ (identified with $\left\{ 0\right\} \times D^1$) forms a second $1$-cell $e^1_2$; $X^1 = e^0 \cup e^1_1 \cup e^1_2$ is homeomorphic to the one-point union of two circles. Finally $D^1 \times D^1$ forms a $2$-cell $e^2$ whose boundary maps to $X^1$ by running over the first circle, then the second, then backwards over the first, then backwards over the second -- the word $aba^{-1}b^{-1} \in \pi_1(S^1 \vee S^1, x)$.

The surface $T^{(g)}$ (genus-$g$ orientable surface) can be thought of as coming from a regular $4g$-gon whose boundary is identified via the pattern $a_1 b_1 a_1^{-1} b_1^{-1} \cdots a_g b_g a_g^{-1} b_g^{-1}$. This gives $T^{(g)}$ the structure of a CW complex with one $0$-cell (from the identified vertices), $2g$ $1$-cells (from the edges, identified in pairs, whose boundaries are sent to the single $0$-cell), and a single $2$-cell coming from the $4g$-gon, whose boundary is mapped into $X^1$ according to the identification pattern; here $X^1$ is homeomorphic to a wedge $W_{2g}$ of $2g$ circles.

Similarly $P^{(h)}$ (the connected sum of $h$ projective planes) has a CW decomposition with a single $0$-cell, $h$ $1$-cells, and a single $2$-cell expressed as a regular $2h$-gon whose boundary is identified according to the pattern $a_1^2 \cdots a_h^2$; here $X^1$ is homeomorphic to the wedge $W_h$ of $h$ circles.

The projective plane is formed from a $2$-disk by identifying $x$ to $-x$ on the boundary circle. It has a CW decomposition $e^0 \cup e^1 \cup_{f_2} e^2$, where $f_2(z) = z^2$. The pseudoprojective plane $P_p$ is formed from the disk by identifying $x$ to $e^{2\pi i/p} x$ on the boundary circle. Give a similar CW decomposition for $P_p$ with one cell in each dimension and identify the attaching map for the $2$-cell.

Consider the CW complex

Here each $1$-cell $e^1_i$ is attached to $X^0$ by identifying its boundary to $e^0_1 \cup e^0_2$. Orient each $1$-cell and identify $\partial e^1_1$ by sending $1$ to $e^0_2$ and $-1$ to $e^0_1$. Similarly, identify $\partial e^1_2$ by sending $1$ to $e^0_1$ and $-1$ to $e^0_2$. To attach the $2$-cells, identify the boundary with $S^1$ and divide it into the upper half arc $S^1_+$ and lower half arc $S^1_-$. Then identify the boundary of $e^2_1$ with $X^1$ by identifying the upper half arc with $e^1_1$ and the lower half arc with $e^1_2$. Identify the boundary of $e^2_2$ using the same map. This is expressed by saying that $\partial e^2_j = e^1_1 + e^1_2$ (see Figure 5.1).

A CW decomposition of the sphere.

More generally, a finite CW complex of dimension $n$ is built up in stages $X^0 \subset X^1 \subset X^2 \subset X^3 \subset \cdots \subset X^n = X$, where $X^i$ is obtained from $X^{i-1}$ by attaching $i$-cells (each an $i$-dimensional disk $e^i$ with boundary $\partial e^i \simeq S^{i-1}$, attached via a continuous map $f : \partial e^i \to X^{i-1}$); a set is open (closed) in $X^i$ iff its inverse image in each cell is open (closed). The subset $X^k$ is called the $k$-skeleton of $X$. Show that the $k$-skeleton $X^k$ is a closed subset of $X$.

Show that a finite CW complex is a compact Hausdorff space.

A subcomplex $Y \subset X$ of a CW complex is a collection of cells of $X$ which is itself a CW complex -- that is, when attaching a cell of $Y$, the image of its attaching map lies within $Y$. Using the CW decomposition of $S^2$ from Exercise 5.1.4, show that the upper hemisphere is a subcomplex.

Show that a subcomplex (Exercise 5.1.7) is a closed subset of $X$.

The $n$-sphere has a CW complex description with one $0$-cell and one $n$-cell, attached by the constant map from $S^{n-1}$ to the $0$-cell. Under the CW topology, the space is the same as taking $D^n$ and identifying all of its boundary points to a single point. Verify that the $n$-sphere is homeomorphic to the space with this CW decomposition.

Give a CW decomposition for $S^1 \times S^2$ by using the product decomposition of the CW decompositions of $S^1$ (one $0$-cell, one $1$-cell) and $S^2$ (one $0$-cell, one $2$-cell), where the product of an $i$-cell and a $j$-cell is homeomorphic to an $(i+j)$-cell. Your description should identify how the skeleta are built up in the product.

Give a CW decomposition for $S^1 \times \mathbb {R} P^2$ by using the product decomposition of the CW decompositions of $S^1$ (one $0$-cell, one $1$-cell) and $\mathbb {R} P^2$ (one cell in each dimension $0$ and $2$, per Exercise 5.1.3 with $p=2$). Your description should identify how the skeleta are built up in the product.

Give a CW decomposition for $S^1 \times S^1 \times S^1$ by using the product decomposition of the CW decompositions of $S^1 \times S^1$ (Exercise 5.1.2(a)) and $S^1$. Your description should identify how the skeleta are built up in the product.

Give a CW decomposition for $S^2 \times S^2$ by using the product decomposition of the CW decomposition of $S^2$ (Exercise 5.1.1). Your description should identify how the skeleta are built up in the product.

Give a CW decomposition for $\mathbb {R} P^2 \times \mathbb {R} P^2$ by using the product decomposition of the CW decomposition of $\mathbb {R} P^2$ (Exercise 5.1.3 with $p=2$). Your description should identify how the skeleta are built up in the product.

Real projective $n$-space $\mathbb {R} P^n$ is formed from the $n$-sphere $S^n$ by identifying antipodal points $x = (x_0, \ldots , x_n) \sim (-x_0, \ldots , -x_n) = -x$. Since $S^n$ has trivial fundamental group for $n \geq 2$, it is the universal cover of $\mathbb {R} P^n$, so $\mathbb {R} P^n$ has fundamental group $\mathbb {Z}_2$. Inductively, going from $\mathbb {R} P^{k-1}$ to $\mathbb {R} P^k$ corresponds to going from $S^{k-1}$ to $S^k$ by attaching the upper and lower hemispheres, each homeomorphic to $D^k$ via vertical projection, attached consistently (the upper hemisphere via vertical projection, the lower hemisphere by composing that attaching map with the antipodal map): $S^k = S^{k-1} \cup_{f_{k-1}} e^k \cup_{a_k f_{k-1}} e^k$, where $f_{k-1}$ is the identity map of $S^{k-1}$ and $a_k$ is the antipodal map of $S^k$. Passing to the quotient identifies the two $k$-cells to give $\mathbb {R} P^k = \mathbb {R} P^{k-1} \cup_{p_{k-1}} e^k$, where $p_{k-1} : S^{k-1} \to \mathbb {R} P^{k-1}$ is the quotient map. This leads to a CW decomposition of $\mathbb {R} P^n$ with $(n+1)$-cells, one in each dimension between $0$ and $n$.

Complex projective $n$-space $\mathbb {C} P^n$, denoted $\mathbb {C} P^n$, is defined analogously: starting from $S^{2n+1} = \left\{ (z_0, \ldots , z_n) \in \mathbb {C}^{n+1} : \sum \left|z_i\right|^2 = 1\right\}$, introduce the equivalence relation $(z_0, \ldots , z_n) \sim \zeta (z_0, \ldots , z_n)$ for $\zeta \in \mathbb {C}$ a unit complex number, and form $\mathbb {C} P^n$ as the quotient $S^{2n+1}/\sim$. For $n = 0$ this identifies all points of the circle to a point, so $\mathbb {C} P^0$ is a point. For $n=1$, one finds $\mathbb {C} P^1 = \mathbb {C} P^0 \cup_{p_0} S^2_+$, where $S^2_+$ is the upper hemisphere of $S^3$ (homeomorphic to $D^2$) and $p_0 : S^1 = \partial S^2_+ \to \mathbb {C} P^0$ is the quotient map -- exactly the CW decomposition of $S^2$ from Exercise 5.1.1, so $\mathbb {C} P^1$ is homeomorphic to $S^2$. The same idea shows that $\mathbb {C} P^k$ is built from $\mathbb {C} P^{k-1}$ by attaching a $2k$-cell.

Fill in the details of the above argument to show that $\mathbb {C} P^n$ has a CW decomposition with one cell in each even dimension between $0$ and $2n$.

Fix an integer $p \geq 2$ and let $\eta = e^{2\pi i / p}$. Let $G_p = \left\{ 1, \eta , \eta^2, \ldots , \eta^{p-1}\right\} \subset S^1$ be the group of $p$th roots of unity, acting on $S^3 = \left\{ (z_0, z_1) : \left|z_0\right|^2 + \left|z_1\right|^2 = 1\right\}$ by $(z_0, z_1) \sim (\tau z_0, \tau z_1)$ for $\tau \in G_p$. The quotient space $L(p,1) = S^3/G_p$ is called a lens space.

We describe the CW decomposition for the case $p = 3$ in detail; the general case follows the same pattern. Sitting inside $S^3$ is the circle $S^1$ where the second coordinate is $0$. Give $S^1$ a CW decomposition consistent with the $G_3$-action: use the points $1, \eta , \eta^2$ as $0$-cells, and $1$-cells $e^1_0, e^1_1, e^1_2$, where $e^1_0$ is mapped via its characteristic map to the arc joining $1$ and $\eta$, and $e^1_1, e^1_2$ are the compositions of this map with multiplication by $\eta , \eta^2$ respectively. Taking the quotient by $G_3$ identifies the $0$-cells to a single $0$-cell and the $1$-cells to a single $1$-cell, and the quotient is homeomorphic to $S^1$.

Next consider the upper hemisphere $S^2_+ \subset S^3$, thought of as a $2$-cell $e^2_0$, attached to $S^1$ so that its boundary is the sum of the three $1$-cells above. Letting $\eta , \eta^2$ act on $e^2_0$ gives two more $2$-cells $e^2_1, e^2_2$ (disjoint from $e^2_0$ except along the shared boundary circle). Taking the quotient by $G_3$ identifies these three $2$-cells, so the quotient has a single $0$-cell, a single $1$-cell, and a single $2$-cell, with the $2$-cell attached so that it runs three times over the $S^1$ formed by the $0$-cell and $1$-cell -- that is, this part of the quotient is $S^1 \cup_{m_3} e^2$ where $m_3(z) = z^3$; this space is the pseudoprojective plane $P_3$ of Exercise 5.1.3.

The $2$-cells $e^2_0$ and $e^2_1$ intersect only along their shared boundary circle, and together (up to homeomorphism) enclose a region homeomorphic to a three-dimensional disk $D^3$, forming a $3$-cell $e^3_0$; all of $S^3$ lies in this $3$-cell or its image under multiplication by $\eta$ or $\eta^2$, giving two more $3$-cells $e^3_1, e^3_2$. Taking the quotient identifies these three $3$-cells, so $L(3,1)$ has a single cell in each dimension between $0$ and $3$.

By replacing $3$ with $p$ throughout the above, give a description of the CW decomposition of the lens space $L(p,1) = S^3/G_p$.

Consider a two-dimensional CW complex $X$ with a single $0$-cell $e^0$, which serves as the base point. Adding the $1$-cells forms a wedge $X^1 = \bigvee_{j=1}^n S^1_j$ of circles, so the attaching map $f : \partial e^2 \to X^1$ of each $2$-cell can be described, up to homotopy (assuming $1$ is mapped to the base point), by a word in the free group $F_n = \pi_1(X^1, e^0)$.

Suppose that there is a single $2$-cell. Use the Seifert--van Kampen theorem to show that the fundamental group of $X$ is isomorphic to the quotient of the free group $F_n$ by the relation given by the word used in attaching the $2$-cell, which is given by $f_*(g)$, where $g$ is the generator of $\pi_1(S^1, 1)$.

In the following exercises (5.2.3-5.2.6), spaces are described as quotients of polygons by indicating identifications of edges; these spaces have natural decompositions as CW complexes, all with a single $0$-cell coming from the vertices. Use these decompositions to compute the fundamental groups, find the abelianizations, and (for those which are surfaces) determine which surface the space is homeomorphic to from the abelianization, checking the result using Euler characteristics.

Find the fundamental groups of the two spaces (a), (b) in Figure 5.2, find their abelianizations, and from this determine which surface each space is homeomorphic to.

Figure for Exercise 5.2.3.

Find the fundamental groups of the two spaces (a), (b) in Figure 5.3 (identifications of the edges of a pentagon, using the same conventions as Exercise 5.2.3).

Figure for Exercise 5.2.4.

Find the fundamental groups of the two spaces (a), (b) in Figure 5.4 (identifications of the edges of a pentagon, using the same conventions as Exercise 5.2.3).

Figure for Exercise 5.2.5.

Find the fundamental groups of the two spaces (a), (b) in Figure 5.5 (identifications of the edges of an octagon, using the same conventions as Exercise 5.2.3).

Figure for Exercise 5.2.6.

Use the Seifert--van Kampen theorem to show that, for a connected CW complex $X$, $\pi_1(X, e^0) \simeq \pi_1(X^2, e^0)$. Thus the fundamental group only detects information about the $2$-skeleton $X^2$.

Use Exercise 5.2.7 to show that $\pi_1(S^k, x) \simeq \left\{ e\right\}$ for $k \geq 3$, since it has a CW decomposition with one $0$-cell and one $k$-cell.

Use the CW decomposition for $S^1 \times S^2$ from Exercise 5.1.10 to compute the fundamental group. Compare your answer with the general result for the fundamental group of a product space.

Use the CW decomposition for $S^1 \times \mathbb {R} P^2$ from Exercise 5.1.11 to compute the fundamental group. Compare your answer with the general result for the fundamental group of a product space.

Use the CW decomposition for $S^1 \times S^1 \times S^1$ from Exercise 5.1.12 to compute the fundamental group. Compare your answer with the general result for the fundamental group of a product space.

Use the CW decomposition for $S^2 \times S^2$ from Exercise 5.1.13 to compute the fundamental group. Compare your answer with the general result for the fundamental group of a product space.

Use the CW decomposition for $\mathbb {R} P^2 \times \mathbb {R} P^2$ from Exercise 5.1.14 to compute the fundamental group. Compare your answer with the general result for the fundamental group of a product space.

Give the fundamental group of $\mathbb {C} P^n$, $n \geq 1$, using its CW decomposition (Exercise 5.1.15).

Show that the fundamental group of the pseudoprojective plane $P_p$ of Exercise 5.1.3 is $\mathbb {Z}_p$.

Show that $S^3$ is the universal covering space of the lens space $L(p,1)$ (Exercise 5.1.16) and use this to give the fundamental group. Then use the CW decomposition from Exercise 5.1.16 to compute the fundamental group in another way.

Let $X$ be a topological space and $f : S^{n-1} \to X$ a continuous function. Let $X_f = X \cup_f D^n$ be the quotient space formed from the disjoint union by identifying $x \in S^{n-1}$ with $f(x) \in X$.

Suppose that $h : X \to Y$ is a homotopy equivalence with homotopy inverse $g : Y \to X$; that is, $gh$ is homotopic to $1_X$ and $hg$ is homotopic to $1_Y$. Name the homotopies $H : X \times I \to X$, $H_t(x) = H(x,t)$, $H_0(x) = gh(x)$, $H_1(x) = x$, and $G : Y \times I \to Y$, $G_t(y) = G(y,t)$, $G_0(y) = hg(y)$, $G_1(y) = y$. Let $X_f = X \cup_f D^n$ (notation of Exercise 5.3.1) for a fixed attaching map $f : S^{n-1} \to X$, and similarly $Y_{hf}$, $X_{ghf}$.

Figure for Exercise 5.3.2.

The dunce hat is the space obtained from a single triangle (a $2$-simplex) by identifying all three of its edges together in the same direction (so that, reading around the boundary of the triangle, all three edges are labeled with the same letter $a$ traversed the same way).

A subcomplex $T$ of a connected $1$-skeleton $X^1$ which contains all of the vertices and deformation-retracts to any one of its $0$-cells (through a process of collapsing $1$-cells one edge at a time) is called a maximal tree in $X^1$. Such a maximal tree can always be shown to exist by induction on the number of $0$-cells: if there is only one $0$-cell, it is itself the maximal tree; otherwise, choosing a $0$-cell $e^0$ and a maximal subcomplex $K$ of $X^1$ not containing $e^0$ (with, say, $p$ path components, each of which has a maximal tree connecting its $0$-cells by the inductive hypothesis), adding a $1$-cell from $e^0$ to each path component of $K$ gives a maximal tree in $X^1$.

Use Exercise 5.3.2 and a maximal tree $T$ in $X$ to show that there is a homotopy-equivalent CW complex $Y$ with a single $0$-cell.

Consider the CW complex whose $1$-skeleton $X^1$ is pictured in Figure 5.10. Here $A, B, C$ denote $0$-cells and $a,b,c,d,e,f$ are $1$-cells. Suppose that there are two $2$-cells attached to $X^1$, which can be described in terms of the $1$-cells that are traversed by $adec$ and $dfb$.

Figure for Exercise 5.3.6.

Consider a handle decomposition of a surface. The $0$-handles deformation-retract to a collection of points. When we attach a $1$-handle, the composition of the attaching maps with the deformation retraction sends these intervals to one or two points. Use this idea to show that the union of the $0$-handles and $1$-handles is homotopy equivalent to a CW complex with $h_0$ $0$-cells and $h_1$ $1$-cells, where $h_i$ is the number of $i$-handles.

Show that if a surface (possibly with boundary) has a handle decomposition with $h_0$ $0$-handles, $h_1$ $1$-handles, and $h_2$ $2$-handles, then there is a corresponding CW complex with $h_0$ $0$-cells, $h_1$ $1$-cells, and $h_2$ $2$-cells to which it is homotopy equivalent.

Show that a connected surface (possibly with boundary) is homotopy equivalent to a CW complex with a single $0$-cell.

For Figure 5.11(a), give a CW decomposition and then get a homotopy-equivalent CW complex with a single $0$-cell. Use this to compute the fundamental group.

Figure for Exercises 5.3.10 and 5.3.11.

For Figure 5.11(b) (shown together with Figure 5.11(a) above, in Exercise 5.3.10), give a CW decomposition and then get a homotopy-equivalent CW complex with a single $0$-cell. Use this to compute the fundamental group.

Using the notation established in Exercise 5.1.4 (where $X$ is the CW complex $e^0_1 \cup e^0_2 \cup e^1_1 \cup e^1_2 \cup_f e^2_1 \cup_g e^2_2$, with each $2$-cell's boundary divided into an upper half arc $S^1_+$ and a lower half arc $S^1_-$, identified respectively with $e^1_1$ and $e^1_2$), consider the CW complex $Y$ where $Y^1 = X^1$ but $Y = X^1 \cup_f e^2_1$, and $\partial e^2$ is identified via sending $S^1_+$ to $e^1_1$ as before but sending $S^1_-$ to $e^1_1$ by reversing the direction. Geometrically, think of $e^1_1$ as $[-1,1]$ on the $x$-axis, with the map being a vertical projection. Identify geometrically what the quotient space will be. Show that $Y$ is homotopy equivalent, but not homeomorphic, to $S^2 \vee S^1$, with the homotopy equivalence coming from collapsing $e^1_1$ to a point.

Suppose $K$ is a CW complex and $L$ is a subcomplex of $K$; then $K$ is built from $L$ by attaching $k$-cells for various $k$, which allows arguments to reduce to the case of attaching a single cell. The homotopy extension theorem states: suppose $L \subset K$ is a subcomplex, $g : K \to Y$, and there is a homotopy of the restriction $g|_L : L \to Y$, that is, a map $H : L \times I \to Y$ with $H(x,0) = g(x)$. Then there is a homotopy $H' : K \times I \to Y$ with $H'(x,t) = H(x,t)$ for $x \in L$ and $H'(x,0) = g(x)$.

To see why this should be true, assume $K = L \cup_f D^k$, where $f : S^{k-1} \to K$ is continuous. The homotopy extension property asserts that a map defined on $L \times I \cup K \times \left\{ 0\right\}$ extends to a continuous map from $K \times I$; the key is to reduce to the case $L = S^{k-1}$, $K = D^k$.

Lemma. Consider $S^{k-1} \times I \cup D^k \times \left\{ 0\right\} \subset D^k \times I$. There is a deformation retraction of $D^k \times I$ onto $S^{k-1} \times I \cup D^k \times \left\{ 0\right\}$.

(Idea of the lemma's proof, illustrated for $k=2$: let $p = (0,2) \in \mathbb {R}^k \times \mathbb {R} = \mathbb {R}^{k+1}$ and consider rays from $p$. For $b \in B = S^{k-1} \times I \cup D^k \times \left\{ 0\right\}$, the ray from $p$ to $b$ meets $D^k \times I$ in an interval with first point on $D^k \times \left\{ 1\right\}$ and last point $b$; for a point $x \in D^k \times I$, moving $x$ along its own such interval to its endpoint $b \in B$ defines the deformation retraction $H$, and $h(x) = H(x,1)$ gives a continuous map $h : D^k \times I \to B$ with $h|_B = \operatorname {id}|_B$.)

Now suppose $g : K = L \cup_f D^k \to Y$ and there is a homotopy $H : L \times I \to Y$ with $H(x,0) = g(x)$ for $x \in L$. Consider the composition $J : S^{k-1} \times I \cup D^k \times \left\{ 0\right\} \to Y$ given by $J(u,t) = H(f(u), t)$ for $u \in S^{k-1}$, and $J(v, 0) = g(v)$ for $v \in D^k$. The lemma allows extending $J$ to $D^k \times I$ by composing with the deformation retraction.

By applying the lemma and using how $K \times I$ is formed as a quotient space, show that there is an extension $H' : K \times I \to Y$ of $H$.

Use an inductive argument, with induction over the number of cells of $K \setminus L$, to prove the homotopy extension theorem (stated in Exercise 5.3.13): if $L \subset K$ is a subcomplex, $g : K \to Y$, and $H : L \times I \to Y$ is a homotopy of $g|_L$ with $H(x,0) = g(x)$, then there is a homotopy $H' : K \times I \to Y$ extending $H$ with $H'(x,0) = g(x)$.

We say that a subset $A \subset X$ is a neighborhood deformation retract if there is an open set $U$ containing $A$ and a deformation retraction of $U$ onto $A$ (a map $h : U \to A$ with $h|_A = \operatorname {id}|_A$ so that $h$ is homotopic to the identity, with the homotopy restricting to the constant homotopy on $A$). If $(A, B)$ is a pair of sets, we say the pair is a neighborhood deformation retract if there are open sets $U, V$ with $A \subset U$, $B \subset V$ so that $U$ deformation-retracts to $A$, $V$ deformation-retracts to $B$, and $U \cap V$ deformation-retracts to $A \cap B$.

Show that if $(A, B)$ is a neighborhood deformation retract of $(U, V)$, then $A, B, A \cap B$ are all path connected iff $U, V, U \cap V$ are all path connected.

Recall (Exercise 5.4.1) that a pair $(A,B)$ is a neighborhood deformation retract of $(U,V)$ if $U, V$ are open sets with $A \subset U$, $B \subset V$, $U$ deformation-retracts to $A$, $V$ deformation-retracts to $B$, and $U \cap V$ deformation-retracts to $A \cap B$. Show that if $A, B, A \cap B$ are path connected and $(A, B)$ is a neighborhood deformation retract of $(U, V)$, then if $x_0 \in A \cap B$, there is an isomorphism between $\pi_1(A, x_0) *_{\pi_1(A \cap B, x_0)} \pi_1(B, x_0)$ and $\pi_1(U, x_0) *_{\pi_1(U \cap V, x_0)} \pi_1(V, x_0)$.