Surjective inclusions in Van Kampen's Theorem

Question

Let $X = U \cup V$ be an open cover. Assume $U,V$ and $U \cap V$ path-connected. We have the inclusions $u : U \cap V \to U$ and $v : U \cap V \to V$ with induced maps $u_* : \pi_1(U \cap V) \to \pi_1 U$ and $v_* : \pi_1 (U \cap V) \to \pi_1 V$. From Van Kampen's Theorem $\pi_1X \cong \pi_1U \ast \pi_1V/N(K)$, where $N(K)$ is the normal closure of $K = \langle u_*(g)v_*(g^{-1})| g \in \pi_1(U \cap V)\rangle$. My Question is:

If $u_*$ and $v_*$ are surjective, is it correct that $$\pi_1X \cong \pi_1(U \cap V)/N_uN_v,$$ where $N_u$ and $N_v$ denote the kernels of $u_*$ and $v_*$ respectively.

From a paper I am reading I understand this to be correct following from van Kampen's theorem and sort of well known. I failed searching the literature and using my bare hands the calculations became too messy very soon.

Answer 1

Yes, this is correct. This is purely group theory, so allow me to rephrase everything solely in group language.

I'll going to set $Y = \pi_1(U\cap V)$. I'll write $U$ for $\pi_1(U)$ and likewise $V$ will denote $\pi_1(V)$.

We have surjective homomorphisms $p_u:Y\rightarrow U$ and $p_v:Y\rightarrow V$. I will, like you, use $N_u$ and $N_v$ for the kernels of these two maps. I will use $N$ to the smallest normal subgroup of $U\ast V$ which contains the set $\{p_u(y) p_v(y^{-1}): y\in Y\}$; so my $N$ is your $N(K)$.

With all this notation, our goal is to prove that $Y/(N_U N_V) \cong (U\ast V)/N$.

We define $\phi: Y/(N_U N_V)\rightarrow (U\ast V)/N$ by $$\phi(y N_U N_V) = p_u(y)N.$$

Further, since $p_u$ is surjective, every element of $U$ has the form $p_u(y)$ for some $y\in Y$. Likewise for $V$. Thus, we can define $\psi:(U\ast V)/N\rightarrow Y/(N_U N_V)$ by $$\psi(p_u(y_1) p_v(y_2) p_u(y_3) p_v(y_4)... N) = y_1 y_2.... N_U N_V.$$

I claim that $\phi$ and $\psi$ are well defined and that they are inverses of each other.

Proposition 1: The map $\phi$ is well defined.

Proof: Let $n_u\in N_U, n_v\in N_V$. We need to show that $\phi(y N_U N_V) = \phi(y n_u n_v N_U N_V)$.

Well, $\phi(y n_u n_v N_U N_V)$ is, by definition, $p_u(y n_u n_v)N$. Since $p_u$ is a homomorphism, and since $u_n \in N_U = \ker p_u$, this is the same as $p_u(y) p_u(n_v)N$.

The element $p_u(n_v) p_v(n_v^{-1}) \in N$ by the definition of $N$. But $n_v\in N_V = \ker p_v$, so $p_v(n_v^{-1})$ is the identity, so $p_u(n_v)\in N$. In other words, $p_u(y)p_u(n_v) N = p_u(y) N = \phi(y N_u N_v)$. Thus, $\phi$ is well defined. $\square$

Proposition 2: The map $\psi$ is well defined.

Proof: Here, there are two things to check. First, that if $g\in U\ast V$ and $n\in N$, that $\psi(g N) = \psi(gn N)$. Second, if $p_u(y_1) = p_u(y_1')$, $p_v(y_2) = p_v(y_2')$, etc, that $\psi(p_u(y_1)p_v(y_2)....N) = \psi(p_u(y_1') p_v(y_2').... N)$.

Let's knock out the first part first. Recall the normal closure of a set is given by arbitrary finite products of conjugates of elements in the set. So, it's enough to show that $\psi(n N)$ is the identity when $n$ is of the form $r p_u(y) p_v(y^{-1}) r^{-1}$ for arbitrary $r\in U\ast V$.

Writing $r = p_u(y_1) p_v(y_2)....$, we get $\psi(r p_u(y) p_v(y^{-1}) r^{-1} N) = y_1 y_2 ... y_n y y^{-1} (y_1 y_2... y_n)^{-1}$ which cancels down to the identity.

Now let's focus on the second issue. If $p_u(y_1) = p_u(y_1')$, etc, then there are $n_1, n_3, n_5,... \in N_u$ and $n_2,n_4,... \in N_v$ with $y_i' = y_i n_i$.

So, we need to show that $y_1 y_2....N_U N_V = y_1 n_1 y_2 n_2 ...N_U N_V$. Here's the idea. Focus on the $n_1 y_2$ on the right side. Rewriting this as $y_2 y_2^{-1} n_1 y_2$, the fact that $N_U$ is normal implies that $y_2^{-1} n_1 y_2 = n_1'$ for some $n_1'\in N_U$. Thus, $n_1 y_2 = y_2 n_1'$.

Repeating this same argument, we can change $y_1 n_1 y_2 n_2....$ into $y_1 y_2 ..... n_1' n_2'......$. Of course, the piece $n_1' n_2'....\in N_U N_V$, so $y_1 y_2... N_U N_V = y_1' y_2'.... N_U N_V$ so $\psi$ is well defined. $\square$

Proposition 3: The maps $\psi$ and $\phi$ are inverses of each other.

Proof: First, $\psi(\phi(y N_u N_v)) = \psi(p_u(y) N) = y N_u N_v$.

Checking the $\phi\circ \psi$ is the identity is a bit messier. It's enough to show that both $\phi(\psi(p_u(y)N)) = p_u(y)N$ and that $\phi(\psi(p_v(y)N)) = p_v(y)N$.

The first is easy: $\phi(\psi(p_u(y)N)) = \phi(y N_u N_v) = p_u(y) N$.

The second gives $\phi(\psi(p_v(y)N)) = \phi( y N_u N_v) = p_u(y) N$, not $p_v(y)N$ like we wanted to see. However, $p_u(y)p_v(y^{-1})\in N$ by definition, so $p_u(y)N = p_v(y)N$ and we are done. $\square$

Answer 2

I dont have a full answer, but I have some progress which might be of interest.

First, there are three maps $\pi_1(U\cap V)\to \pi_1(X)$:

1. $i_*$ corresponding to the inclusion $i:U\cap V\subset X$.
2. By Van Kampen, the composition $\pi_1(U\cap V)\to \pi_1(U)\to \pi_1(U)*\pi_1(V)\to \pi_1(X)$
3. Similarly for $V$

The Van Kampen theorem gives us that all 3 maps are actually the same, and that this map is surjective. Now we need to see that $N_u,N_v$ maps to $0$ and that their product is the entire kernel. The first part follows by choosing either definition 2 or 3 for the map.

Answer 3

I will present a solution that allows us to avoid using representations of elements in the kernel $K$.

Suppose $X=U \cup V$ where $U$ and $V$ satisfy your hypothesis, and the maps $u_*$ and $v_*$ and $N_u$ and $N_v$ and $K$ are as described above.

Define a homomorphism $\varphi: \pi_1(A \cap B)/N_uN_v \rightarrow (\pi_1(U) * \pi_1(V))/K$ by

$gN_uN_v \mapsto u_*(g)K$.

Observation 1: $u_*(g)K=v_*(g)K$ by definition of K.

Observation 2: $\varphi$ is well defined.

Since if $gN_uN_v=g'N_uN_v$ then $g=g'n_un_v$ for some $n_u \in N_u$ and $n_v \in N_v$. Then $\varphi(gN_uN_v)=\varphi(g'n_un_vN_uN_v)=u_*(g'n_un_v)K=u_*(g')u_*(n_u)u_*(n_v)K=u_*(g')u_*(n_u)v_*(n_v)K=u_*(g')K$

Now we will to construct an inverse homomorphism for $\varphi$.

Observation 3: Since $u_*; \pi_{1}(U \cap V) \rightarrow \pi_1(U)$ is surjective, it induces an isomorphism $\bar{u}:\pi_1(U \cap V)/N_u \rightarrow \pi_1(U)$ and, similarly, $v_*$ induces an isomorphism $\bar{v}:\pi_1(U \cap V)/N_v \rightarrow \pi_1(V)$.

Note that, by definition, $\bar{u}(gN_u)=u_*(g)$ and $\bar{v}$ is defined similarly.

Now define $j_u: \pi_1(U \cap V)/N_u \rightarrow \pi_1(U \cap V)/N_uN_v$ by $j_u(gN_u)=gN_uN_v$ and similarly define $j_v: \pi_1(U \cap V)/N_v \rightarrow \pi_1(U \cap V)/N_uN_v$

Observation 4: $j_u$ and $j_v$ are well defined. This is easy to check.

Define $\mu=j_u \circ \bar{u}^{-1}$ and $\nu=j_v \circ \bar{v}^{-1}$. By the universal property of free products, $\mu$ and $\nu$ extend to a homomorphism

$\psi: \pi_1(U) * \pi_1(V) \rightarrow \pi_1(U \cap V)/N_uN_v$.

Observation 5: $K \subset \ker \psi$.

This is also easy to check using the fact that $\bar{u}^{-1}(u_*(g))=g$ and similarly $\bar{v}^{-1}=(v_*(g))=g$

Since $K \subset \ker \psi$, we know that $\psi$ induces a homomorphism

$\bar{\psi}:(\pi_1(U)*\pi_1(V))/K \rightarrow \pi_{1}(U \cap V)/N_uN_v$

defined by $\bar{\psi}(gN)=\psi(g)$.

Observation 6: $\bar{\psi}$ is the inverse of $\varphi$. This follows straight from the definitions easily.

Conclude that $\varphi$ is an isomorphism.