Jae Choon Cha

POSTECH

### Pushout

In this lecture, all diagrams are assumed to be commutative, unless explicitly stated otherwise.

Consider the following diagram of groups and homomorphisms.

It is called a *pushout diagram* if for every commutative diagram of the form

there is a unique homomorphism $\phi\colon G\to H$ satisfying $\phi \alpha = \gamma$ and $\phi\beta=\delta$. See the following diagram which combines the above two:

If it is the case, the group $G$ is called a *pushout* of the following
diagram.

The above defining property of a pushout diagram is called the *universal
property* of the pushout.

A standard argument using the universal property of $G$ proves that a pushout is unique up to isomorphism if it exists. Moreover, the following theorem says that the pushout always exists (for the case of groups), and it is easy to compute when we have group presentations.

**Theorem.** Suppose $A$, $B$ and $C$ are given as the following group
presentations:

For given homomorphisms $f\colon C\to A$ and $g\colon C\to B$, Let

Define $\alpha\colon A\to G$ and $\beta\colon B\to G$ by $\alpha(x_i)=x_i$, $\beta(y_i)=y_i$. Then the following is a pushout diagram.

*Proof.* First, note that $\alpha$ and $\beta$ are well-defined homomorphisms by
the property of a group presentation, since the association sends relators to
relators.

To show the universal property of the pushout, suppose $\gamma\colon A \to H$ and $\delta\colon B\to H$ are homomorphisms satisfying $\gamma f = \delta g$. Define $\phi\colon G \to H$ by $\phi(x_i) = \gamma(x_i)$, $\phi(y_i) = \delta(y_i)$. Note that $\phi(r_i)=\gamma(r_i)$ and $\phi(s_j) = \delta(s_j)$ are trivial in $H$. Furthermore, $\phi(f(z_k)g(z_k)^{-1}) = \gamma(f(z_1))\delta(g(z_1))^{-1}$ is trivial since $\gamma f = \delta g$. It follows that $\phi$ is a well-defined homomorphism, once again by the property of a group presentation. The homomorphism $\phi$ satisfies $\phi\alpha=\gamma$ and $\phi\beta=\delta$ by its definition.

If $\psi$ is another homomorphism satisfying $\psi\alpha=\gamma$ and $\psi\beta=\delta$, then we must have $\psi(\alpha(x_i)) = \psi(x_i) = \gamma(x_i)$ and $\psi(\beta(y_i)) = \psi(y_i)=\delta(y_i)$. So, $\psi=\phi$ on each $x_i$ and $y_i$, and thus $\psi=\phi$. This shows the required uniqueness. $\quad\square$

### Statement of the Seifert-van Kampen theorem

**Theorem.** Suppose $U$ and $V$ are open subsets of a space $X$ such that
$U\cap V$ is path connected and $U\cup V = X$. Fix a point $x_0\in U\cap V$,
and let $i$, $j$, $k$ and $\ell$ be the inclusions in the following diagram.

Then, the following diagram of induced group homomorphisms is a pushout diagram.

The theorem could be understood as follows. Observe that the notion of pushout
which was defined about does not use anything specific to groups, except that we
said the arrows are group homomorphisms. Indeed, it is possible to define the
notion of pushout for other kinds of objects. Indeed, when you learn the notion
of categories, it will be formally generalized in a more abstract way. Even for
now, it is straightforward to formulate the pushout for spaces and continuous
maps. Namely, just replace groups with spaces and homomorphisms with maps, to
define a pushout diagram of spaces. Then, it is easily verified that the above
diagram consisting of $U\cap V$, $U$, $V$ and $X$ is a pushout diagram. So, our
theorem says: *$\pi_1(-)$ takes a pushout diagram to a pushout diagram* (when
$U$, $V$ satisfies the hypothesis of the theorem).

### Examples

We first discuss some examples, postponing the proof. An extremely useful feature of the Seifert-van Kampen theorem is that when the fundamental groups of $U\cap V$, $U$ and $V$ are given as group presentations, it is very easy to compute a group presentation of the fundamental group of $X=U\cup V$, using the above algebraic theorem on the pushout presentation.

#### Spheres of dimension${}\ge 2$

We will show that $\pi_1(S^n,*)$ is trivial for $n\ge 2$. (Since $S^n$ is path connected, the conclusion is irrevant to the choice of a basepoint $*$.) To prove this, let $U=S^n\sm \{p\}$ and $V=S^n\sm \{q\}$, where $p$ and $q$ are the north and south pole of the sphere $S^n$. Then $U$ and $V$ are open, and $U\cap V$ is path connected. (For this path connectedness, it is essential to assume $n\ge 2$.) So the Seifer-van Kampen theorem applies to our situation, to give the following pushout diagram.

Observe that $S^n\sm \{p\}$ is homeomorphic to $\R^n$ (eg. via the streographic projection) which is contractible. Therefore $\pi_1(S^n\sm \{p\},*)$ is trivial. Similarly $\pi_1(S^n\sm \{q\},*)$ is trivial too. Now, compute the presentation of the pushout $\pi_1(S^n,*)$, using the empty presentation $\langle \cdot \mid \cdot \rangle$ for $\pi_1(S^n\sm \{p\},*)$ and $\pi_1(S^n\sm \{p\},*)$. This gives us that $\pi_1(S^n,*)$ has an empty presentation.

#### Two-loop graph

Consider the graph $X$ consisting of one vertex and two edges:

Let $U$ and $V$ be the open subsets shown in the diagram. (More precisely, $U$ is the relative open set in $X$ obtained by taking the intersection of $X$ and the interior of the dashed open ellipse in the plane, and $V$ is given similarly.) Look at the pushout diagram given by the Seifert-van Kampen theorem:

In our case, $U$ and $V$ are homotopy equivalent to a circle, and thus we have $\pi_1(U,*)=\langle x\mid \cdot \rangle$ and $\pi_1(V,*)=\langle y\mid \cdot \rangle$. Here $x$ and $y$ are the loops given by the edges $x$ and $y$. (Abuse of notation!) Also, $U\cap V$ is homeomorphic to the union of the $x$-axis and $y$-axis in the plane, and thus $U\cap V$ is contractible and $\pi_1(U\cap V,*) = \langle \cdot \mid \cdot \rangle$.

Therefore, it follows that

Namely, $\pi_1(X,*)$ is the free group of rank 2, generated by $x$ and $y$.

#### Real projective plane

The real projective plane $\R P^2$ is defined to be the space of all lines in $\R^3$ passing through the origin. To make it a topological space, the following reinterpretation is useful:

Namely, $\R P^2$ is defined to be the above quotient space of the topological space $\R^3\sm \{0\}$. It is straightforward to see the following:

where $f$ in the last line is the map $f\colon \partial D^2 \to S^1$ given by $f(z)=z^2$.

Using this, view $\R P^2$ as a quotient space of the disjoint union $D^2\cup S^1$, and let $q\colon D^2\cup S^1 \to \R P^2$ be the quotient map. Here, although $\partial D^2=S^1$, we will always denote it by $\partial D^2$ in order to avoid confusion with $S^1$ in $D^2\cup S^1$. Fix $*=q(\frac12)$ (the image of $\frac12 \in D^2$) as the basepoint. To compute $\pi_1(\R P^2)$, let $U=q(\operatorname{int} D^2)$ and $V=q(D^2\sm \{0\})$. $U$ and $V$ are open subsets in $\R P^2$, and $U\cup V=\R P^2$. Also, $U\cap V = q(\operatorname{int}D^2 \sm \{0\})\cong \operatorname{int}D^2 \sm \{0\}$ is path connected.

Consider the following pushout diagram:

Since $U \cong \operatorname{int} D^2$ is contractible, $\pi_1(U,*)$ is trivial. Since $V\simeq q(\partial D^2) \cong S^1$, $\pi_1(V,*)=\langle x\mid \cdot\rangle$ where $x$ is the generator of $\pi_1(S^1)$. Since

we have $\pi_1(U\cap V, *) = \langle z \mid \cdot \rangle$. Here, $z$ is the loop in $q(\operatorname{int} D^2\sm \{0\})$ given by $s\mapsto q(\frac12 e^{2\pi i s})$. Since the image of $z$ in $\pi_1(V,*)$ is $x^2$, we have the following:

### Proof of the Seifert-van Kampen theorem

Suppose $U$, $V$, and $X$ are as in the statement of the theorem. Our goal is to prove that

is a pushout diagram. For this purpose, we will verify the universal property. Suppose $f\colon \pi_1(U,x_0) \to H$ and $g\colon \pi_1(V,x_0)$ are homomorphisms satisfying $fi_* = gj_*$. It suffices to prove that there is a unique homomorphism $\phi\colon \pi_1(X,x_0) \to H$ satisfying $\phi k_* = f$ and $\phi \ell_* = g$.

To define $\phi$, let $\alpha\colon I \to X$ be a loop based at $x_0$. Since $\{U,V\}$ is an open cover for $X$, there is a subdivision $0=s_0 < s_1 < \cdots < s_n=1$ such that $\alpha([s_{i-1},s_{i}])$ is contained in either $U$ or $V$ for each $i$. We may assume that $\alpha(s_i) \in U\cap V$, by deleting some $s_i$ if necessary. Let $\alpha_i\colon I \to X$ be the path $\alpha_i(s) = \alpha((1-s)\cdot s_{i-1} + s\cdot s_{i})$ for $i=1,\ldots,n$, that is, $\alpha_i$ is a reparametrization of $\alpha|_ {[s_{i-1},s_i]}$. Each $\alpha_i$ lies in $U\cap V$, but they are not loops in general. To produce loops from the $\alpha_i$, we proceed as follows. For $i=1,\ldots, n-1$, choose a path $\gamma_i$ in $U\cap V$ from $x_0$ to $\alpha(s_i)$. (Here we use the path connectedness of $U\cap V$!) For $i=0$ and $n$, let $\gamma_0=c_{x_0}$, $\gamma_n = c_{x_1}$. Now, for $i=1,\ldots,n$, the product $\gamma_{i-1}^{\vphantom{1}}\alpha_i^{\vphantom{1}}\gamma_i^{-1}$, which is well-defined, is a loop based at $x_0$ and satisfies the following:

Moreover, $\gamma_{i-1}^{\vphantom{1}}\alpha_i^{\vphantom{1}}\gamma_i^{-1}$ is a loop contained in either $U$ or $V$. Define, for $i=1,\ldots,n$,

Now define $h_\alpha = h_1\cdots h_n \in H$.

**Claim 1.** $h_\alpha$ is independent of the choice of the paths $\gamma_i$.

*Proof of Claim.* Suppose $\delta_i$ is another path in $U\cap V$ from $x_0$ to
$\alpha(s_i)$ for $i=1,\ldots,n-1$, and $\delta_0=c_{x_0} = \delta_n$. Define
$k_i \in H$ similarly to $h_i$, but now using $\delta_i$ in place of $\gamma_i$:

Then we have

when $\alpha_i$ lies in $U$. Similarly, when it lies in $V$, we have

Since $[\gamma_{i-1}^{\vphantom{1}} \delta_{i-1}^{-1}] \in \pi_1(U\cap V, x_0)$, we have $f([\gamma_{i}^{\vphantom{1}} \delta_{i}^{-1}]) = g([\gamma_{i}^{\vphantom{1}} \delta_{i}^{-1}])$. From this it follows that

by cancellation. This shows that $h_\alpha$ is independent of the choice of $\gamma_i$.

**Claim 2.** $h_\alpha$ is independent of the choice of the subdivision $0=s_0 < s_1 < \cdots < s_n=1$.

*Proof of Claim.* Since any two subdivisions have a common refinement, it
suffices to show the claim when we adjoin an additional point to a given
subdivision $0=s_0 < s_1 < \cdots < s_n=1$. Suppose $r\in (s_{i-1},s_i)$ is a
point such that $\alpha(r)\in U\cap V$. Choose $\delta$ from $x_0$ to
$\alpha(r)$. Let $\beta(s)=\alpha((1-s)\cdot s_{i-1} + s\cdot r)$, $\beta'(s) = \alpha((1-s)\cdot r + s\cdot s_i)$. That is, $\beta$ and $\beta'$ are
reparametrized $\alpha|_{[s_{i-1},r]}$ and $\alpha|_{[r,s_{i}]}$. If
$\alpha_i$ is in $U$, then we have

Also, a similar identity holds when $\alpha_i$ is in $V$. It follows that $h_\alpha$ remains the same when $r$ is added.

**Claim 3.** $h_\alpha$ is invariant under homotopy of $\alpha$ rel $\{0,1\}$.

*Proof of Claim.* Suppose $F\colon I\times I \to X$ is a homotopy between two
loops based at $x_0$ rel $\{0,1\}$. By Lebesgue’s lemma, there is $\epsilon>0$
such that $F([s,s+\epsilon]\times[t,t+\epsilon])$ is contained in either $U$ or
$V$. Fix $t_0, t_1 \in I$ such that $0 < t_0 – t_1 < \epsilon$. It suffices to
show that $h_\alpha = h_\beta$ for the two loops $\alpha(s)=F(s,t_0)$ and
$\beta(s)=F(s,t_1)$. Choose $0=s_0 < s_1 < \cdots < s_n=1$ such that
$F([s_{i-1},s_i] \times[t_0,t_1]) \subset U$ or $V$ (e.g., it holds if
$s_i-s_{i-1} < \epsilon$). May assume $F(\{s_i\}\times[t_0,t_1])\subset U\cap V$. Choose paths $\gamma_i$ in $U\cap V$ from $x_0$ to $\alpha(s_i) = F(s_i,t_0)$ for $i=1,\ldots,n-1$ and let $\gamma_0=\gamma_n = c_{x_0}$. Let
$u_i(s)=F(s_i, (1-s)\cdot t_0 + s\cdot t_1)$ and let $\delta_i = \gamma_i u_i$.
Consider $h_\alpha$ and $h_\beta$ defined using $\gamma_i$ and $\delta_i$. Then

and therefore

From this it follows that $h_\alpha=h_\beta$. This shows the claim.

Now, define $\phi\colon \pi_1(X,x_0) \to H$ by $\phi([\alpha]) = h_\alpha$. By the above claims, $\phi$ is a well-defined function. Moreover, from the definition of $h_\alpha$ it easily follows that $h_{\alpha}h_{\beta} = h_{\alpha\beta}$. That is, $\phi$ is a homomorphism. If $\alpha$ is a loop in $U$ based at $x_0$, then $h_\alpha=f[\alpha]$ by using the trivial subdivision $0=s_0 < s_1 = 1$, and thus $\phi(k_*[\alpha]) = f[\alpha]$. Similarly $\phi(\ell_*[\alpha]) = g[\alpha]$. So $\phi$ makes the desired diagram commutative.

For the uniqueness, suppose $\psi\colon \pi_1(X,x_0)\to H$ is another homomorphism satisfying $\psi k_* = f$ and $\psi \ell_* = g$. Then for

we have

when $\alpha_i$ is contained in $U$. Simiarly, $\psi[\gamma_{i-1}^\mathstrut \alpha_{i\vphantom{1}}^\mathstrut\gamma_{i\vphantom{1}}^{-1}] = \phi[\gamma_{i-1}^\mathstrut \alpha_{i\vphantom{1}}^\mathstrut\gamma_{i\vphantom{1}}^{-1}]$ when $\alpha_i$ is contained in $V$. It follows that $\psi[\alpha] = \phi[\alpha]$. This shows the uniqueness, and completes the proof of the Seifert-van Kampen theorem.