# Open Sets in the Wedge Sum and a Homeomorphism

I am presently working through example 1.21 in Hatcher's book on wedge sums of topological spaces. He makes a few claims which I am having trouble verifying. First, let me set-up some notation.

Let $$\{X_i\}_{i \in I}$$ be a collection of topological spaces. Then $$\amalg_{i \in I} X_i := \cup_{i \in I} \{(x,i) \mid x \in X_i\}$$ is the disjoint union of the topological spaces, endowed with the finest topology with respect to which every map $$\varphi_i : X_i \to \amalg_{i \in I} X_i$$ defined by $$\varphi_i(x) = (x,i)$$ is continuous. It is clear that $$\amalg_{i \in I} X_i = \cup_{i \in I} \varphi_i(X_i)$$.

Let $$x_i \in X_i$$. Then the wedge sum of these topological spaces is defined as $$\vee_{i \in I} X_i := \amalg_{i \in I} X_i/ \sim$$ with the equivalence relation $$\sim$$ defined such that all points $$(x_i,i)$$ are equivalent/identified (i.e., they all lie in the same equivalence class). Letting $$p : \amalg_{i \in I} X_i \to \vee_{i \in I} X_i$$ denote the canonical quotient map generating the quotient topology, it is clear that $$\vee_{i \in I} X_i = \cup_{i \in I} p(\varphi_i(X_i))$$

In example 1.21, Hatcher says

"If each $$x_i$$ is a deformation retract of an open neighborhood $$U_i$$ in $$X_i$$, then $$X_i$$ is a deformation retract of its open neighborhood $$A_i = X_i \vee_{i \neq j} U_j$$.

First, is "$$X_i \vee_{i \neq j} U_j$$" a misprint; should it actually be $$X_i \vee \bigvee_{j \neq i} U_j$$? Second, strictly speaking $$X_i$$ is not a subset of the wedge sum. But I suspect he means that $$X_i$$ is homeomorphic to some subspace of the wedge sum that deformation retracts onto $$A_i$$. My first guess was that $$X_i$$ is homeomorphic to $$p(\varphi_i(X_i))$$. Certainly $$p \varphi_i$$ is a continuous surjection from $$X_i$$ to $$p(\varphi_i(X_i))$$; it is relatively simply to show that it is injective; and $$\varphi_i$$ is an open map. I tried showing that $$p$$ is an open map, but I couldn't figure it out...So it isn't clear how to show that $$X_i$$ is homeomorphic to $$p( \varphi_i(X_i))$$ (perhaps it isn't).

Let $$U \subseteq \amalg_{i \in I} X_i$$ be open. Then $$p(U)$$ is open if and only if $$p^{-1}(p(U))$$ is open. If $$U$$ does not contain any of the points that get glued together, then I believe $$p^{-1}(p(U)) = U$$, so $$p(U)$$ is open. However, if it contains at least one point, then $$p^{-1}(p(U)) = U \cup \{(x_i,i) \mid i \in I\}$$. If I could show that $$\{(x_i,i) \mid i \in I\}$$ is open, then I'd be finished...but I have a feeling it isn't...

Third, is the wedge sum of open subsets open in $$\bigvee_{i \in I} X_i$$? If I could show that $$p$$ is open, then I believe it is obviously true.

• Surely you jest. Last time I checked, equality was symmetry. How is $i\ne j$ different from $j\ne i$? – Ted Shifrin Apr 30 at 0:36

Maybe you would be best served by a simple example, $$S_1 \approx S_2 \approx S^1$$ with points $$s_i\in S_i$$ and $$X=S_1 \vee S_2$$. Now what would I mean by $$S_1 \subset X?$$ Well there is a bijective correspondence to the equivalence class of points in $$X$$ which is a quotient, namely for $$s\in S_1$$ with $$s\neq s_1$$ the quotient has no effect on $$(s,1)\in S_1 \amalg S_2$$ and for $$s=s_1$$ we have the identification $$s\mapsto (s_1,1)\sim (s_2,2)$$.
The reason for this particular example is we can make a picture (figure 8) and have no qualms about labeling the two loops $$S_1$$ and $$S_2$$. Now what are the basic open sets in $$S_1 \subset X$$. Well they are the intersection of basic open sets of $$X$$. That is they are open intervals contained on our loop labeled $$S_1$$ not containing $$s_1$$ or they are open crosses at the wedge so intersect that with $$S_1$$ and we have an open interval on $$S_1$$ containing $$s_1$$. There is our correspondence between basic open sets; these spaces really are homeomorphic and the notation $$S_1 \subset X$$ is very natural.
To answer the very first point $$X_i \vee_{i\neq j} U_j$$ is fine, as $$x_j \in U_j$$ be construction the wedge there makes perfect sense.