# Surjective inclusions in Van Kampen's Theorem

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.

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. $$\square$$

• This is easily one of the least elegant answers I've every worked out on MSE. I'd LOVE to see a slicker argument! Apologize to anyone who slogged through the whole thing! Commented Feb 12, 2019 at 6:11
• Apologies*, not apologize! Commented Feb 12, 2019 at 13:43
• I would like to see an elegant solution also.
– Tuo
Commented Feb 13, 2019 at 0:26
• Cool, thanks! It was quite enjoyable getting back to this after not having done proper mathematics in two years Commented Feb 14, 2019 at 10:19

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.

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.