2
$\begingroup$

I want to prove the following result:

Let $X, Y$ be topological spaces and $X\vee Y$ it's wedge sum on $p$. If there exist simply connected neighborhoods of $p$ on $X$ and $Y$ then $\pi_1(X\vee Y, p) = \pi_1(X,p) * \pi_1(y,p)$

My attempt:

If we call $A$ and $B$ to such neighborhoods on $X$ and $Y$ respectively, then using $A,B,X\cup B, Y\cup A$ we get open subsets where we can use Van Kampen's theorem.

Because $A$ and $B$ are simply connected their fundamental groups are trivial.

The problem is to find the $\pi_1(X\cup B, p)$ and $\pi_1(Y\cup A, p)$. Because as $X$ is not open on the wedge sum I can't use Van Kampen using that set.

I've seen on Hatcher's Algebraic Topology that for calculating $\pi_1(S^1 \vee S^2)$ uses $\pi_1(S^1)$ and $\pi_1(S^2)$ but I don't see why I can do this if these sets are not open on wedge sum.

$\endgroup$
5
  • 2
    $\begingroup$ This is false, and the usual counterexample is the wedge of two copies of the cone on the Hawaiian earring: each is simply connected but the wedge is not. You need to demand that in one of the spaces, there is a neighborhood of the basepoint which deformation retracts onto that basepoint. $\endgroup$
    – user98602
    Commented Dec 16, 2018 at 0:52
  • $\begingroup$ Oh sorry that’s what i ment. Aldeady added that the neighborhood os of $p$ $\endgroup$
    – Johanna
    Commented Dec 16, 2018 at 1:08
  • 2
    $\begingroup$ The point is that "simply connected neighborhood" is not sufficient. In the above example, $X = Y$ is the cone on the Hawaiian earring and $A$ is just the whole of $X$. $\endgroup$
    – user98602
    Commented Dec 16, 2018 at 1:15
  • $\begingroup$ The result you're trying to prove is a special case of the Seifert-van Kampen theorem. $\endgroup$
    – freakish
    Commented Dec 16, 2018 at 12:09
  • $\begingroup$ @freakish That is not true! The post here is precisely about addressing a subtlety in this application: SvK uses open sets and our input is the wedge summands of $X \vee Y$, which are not open. Choosing the appropriate open sets is the important content here. $\endgroup$
    – user98602
    Commented Dec 19, 2018 at 9:04

1 Answer 1

3
$\begingroup$

Disclaimer: as stated this result is not true, as exemplified in the comments. I'm not very knowledgeable in the subject matter, but having thought about this problem recently, maybe I can offer you some intuition as to why this fails, and a more restrictive case in which it works. $$ \\ \rule{10cm}{0.5pt} \\ $$ Let's first look at what an open set of $X \vee Y$ is like. Since $X \vee Y$ is defined as $X \sqcup Y / \sim$ with $\sim$ the identification of certain points $x_0 \in X$, $y_0 \in Y$ (which we note $p := [x_0] = [y_0]$), endowed with the final topology, $U \subset X \vee Y$ will be open only if it's preimage via the projection $q : X \sqcup Y \to X \vee Y$ is open. Open sets of a disjoint union are union of open sets of each factor, and we have the additional condition that if the open set of $X$ contains $x_0$, the open set of $Y$ ought to contain $y_0$ and viceversa. This is simply because, for example, $x_0 \in q^{-1}(U)$ implies $p = [y_0] \in U$ and so $y_0$ is also on the preimage of $U$. Hence the neighbourhoods of $p$ on the wedge sum are of the form $q(A_X \sqcup A_Y)$ with $x_0 \in A_X \subset X$ and $y_0 \in A_Y \subset Y$.

To use van Kampen, one would like to have two suitable open sets of $X \vee Y$. This implies two choices of open sets for each space. A reasonable option could be $X \cup U$ and $V \cup Y$ with $U$ and $V$ the image of the neighbourhoods given. Then $(X \cup U) \cap (Y \cup V) = U \cup V$ and thus

$$ \pi_1(X \vee Y) \simeq \frac{\pi_1(X \cup U) * \pi_1(Y \cup V)}{\pi_1(U \cup V)}. \tag{1} $$

and so one would hope for $U \cap V$ to be simply connected and the other two homotopic to $X$ and $Y$ respectively. Intuitively it may seem clear, but there is a catch (in all cases): for example, when trying to deform $X \cup U$ into $X$, even if $U$ were contractible there is not a guarantee that the homotopy from $U$ to $p$ will fix $p$ at each instant. Thus, it is unreasonable to expect for this map to factor through the quotient, i.e. to induce a homotopy from $X \cup U$ to a point, because 'moving $p$ would imply moving $X$ as well', something we have no control of.

Indeed, as Mike Miller explains in the comments, it is not enough for the neighbourhoods to be contractible (let alone simply connected). One possible counterexample seems to be the space $\mathbb{G} := CH \vee CH$, the wedge of two hawaiian earrings $H$ at the 'trouble point'. This article seems to have a construction of it as well as a discussion on why our claim fails. Just looking at the pictures is at least a bit helpful in order to readjust our intution. The idea is that since cones are always contractible, the gluing point has each space as a contractible neighbourhood. However, when gluing we fail to be able to collapse one cone without moving the other, and as a consequence the fundamental group is not trivial,

$$ \pi_1(CH) * \pi_1(CH) = 0 * 0 = 0 \neq \pi_1(CH \vee CH) = \pi_1(\mathbb{G}). $$

A positive result can be obtained, however, if we assume that $p$ is a deformation retract of both neighbourhoods (it is claimed in the comments that we only need one neighbourhood to fulfill this condition, but the proof I am aware of uses this fact for both of them). We proceed as we did in $(1)$, but this time we effectively get that $U \cup V \simeq *$, $X \cup U \simeq X$ and $Y \cup V \simeq Y$ as special cases of the following result:

Proposition. Let $X \vee Y$ be the wedge of two spaces at some point $p = [x_0] = [y_0]$ with $x_0 \in X$, $y_0 \in Y$, and suppose that $\{y_0\}$ is a strong deformation retract of $Y$. Then, $X \vee Y \simeq X$.

Proof. Let $H : 1_Y \simeq y_0$ be a homotopy from the identity on $Y$ to the constant function at $y_0$, such that for all $t\in I := [0,1]$ we have that $H_t(y_0) := H(y_0,t) = y_0$. Now, a map on $X \sqcup Y$ is continuous if and only if it is continuous restricted to $X$ and $Y$, and so

$$ \begin{align} K :& (X \sqcup Y) \times I \rightarrow X \sqcup Y\\ & (z,t) \longmapsto \cases{z \qquad \ \ \text{ if $x \in X$} \\ H(y,t) \text{ if $y \in Y$}} \end{align} $$

is continuous. Composing with $q \times 1_I$ for $q : X \sqcup Y \to X \vee Y$ the projection, we get a continuous function

$$ \begin{align} \tilde{K} :& (X \sqcup Y) \times I \rightarrow X \vee Y\\ & (z,t) \longmapsto \cases{[z] \qquad \ \ \text{ if $x \in X$} \\ [H(y,t)] \text{ if $y \in Y$}} \end{align} $$

and since now $H(x_0,t) = [x_0] = p = [y_0] = H(y_0,t)$ for all $t \in I$, then $\tilde{K}$ induces a map

$$ \begin{align} L :& (X \vee Y) \times I \rightarrow X \vee Y\\ & ([z],t) \longmapsto \cases{[z] \qquad \ \ \text{ if $x \in X$} \\ [H(y,t)] \text{ if $y \in Y$}} \end{align} $$

Now, take the inclusion $i : X \hookrightarrow X \vee Y$ and $g : X \vee Y \to X$ with $g([x]) = x$ and $g([y]) = x_0$ for $x \in X, y \in Y$, which 'collapses $Y$'. It is clear that $gi = 1_X$, and $ig \simeq 1_{X \vee Y}$ via the function $L$ we have previosuly constructed: this is to say that $X$ and $X \vee Y$ are homotopic. $\square$

As a side note, the homotopy $L$ fixes $X$. Hence we get a stronger conclusion for free: with these hypotheses, $X$ is actually a strong deformation retract of $X \vee Y$.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks that was really helpful! $\endgroup$
    – Johanna
    Commented Dec 16, 2018 at 23:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .