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. 
 A: 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.
