What does 'copy' mean in topology?

Question

I don't know the meaning of 'copy' in the following exercise (Munkres. "Topology" 2/e. p. 370. exercise 59.1.):

Let $X$ be the union of two copies of $S^2$ having a point in common. What is the fundamental group of $X$? Prove that your answer is correct. [Be careful! The union of two simply connected spaces having a point in common is not necessarily simply connected.]

Does it mean a space which is homeomorphic to $S^2$? If I'm right, then how is the topology of $X$(a union two copies) determined?

Answer 1

Usually, a copy of a topological space $Y$ is a topological space $Y'$ which is homeomorphic to $Y$, but distinct from it in the sense that $Y\neq Y'$.

What the question is describing is basically the space

$$S((-1,0,0), 1)\cup S((1,0,0), 1)$$

where $S(x, r)$ is the sphere, centered around $x$ and with a radius of $r$ (i.e. it's two spheres touching on one point).

Answer 2

Let $X$ be the union of two copies of $S^2$ having a point in common.

This is a very non-formal (and not many mathematicians would use that wording) way of simply saying that $X=S^2\vee S^2$ where $\vee$ is the wedge sum operator a.k.a. glueing at a point. Or in other words these are two spheres touching each other at exactly one point:

As you can see from the image these are two "copies" of $S^2$.

Answer 3

Munkres is a little imprecise when defining $X$ as the union of two copies of $S^2$ having a point in common. In fact, details are left to the reader's imagination.

There is a general construction of the "one-point union" or "wedge" of two spaces $X_i$ endowed with base points $x_i^0$ ($i = 1, 2$). Let us denote the pair $(X_i,x_i^0)$ as a pointed space. Then we can construct a pointed space $(W,w^0) = (X_1,x_2^0) \vee (X_2,x_2^0)$ and two continuous base point preserving maps $j_i : (X_i,x_i^0) \to (W,w^0)$ with the following universal property:

For any two continuous base point preserving maps $f_i : (X_i,x_i^0) \to (Z,z^0)$ to any poined space $(Z,z^0)$ there exists a unique continuous base point preserving map $F : (W,w^0) \to (Z,z^0)$ such that $F \circ j_i = f_i$.

If you know some category theory, you will recognize this construct as the sum in the category of pointed spaces and continuous base point preserving maps.

There are various explicit constructions. They are all equivalent in the sense that if $(\tilde{W},\tilde{w}^0), \tilde{j}_1, \tilde{j}_2$ form another system with the above universal property, then there exists a unique base point preserving homeomomorphism $h : (W,w^0) \to (\tilde{W},\tilde{w}^0)$ such that $h \circ j_i = \tilde{j}_i$.

Here are three explicit constructions (verification of the universal property omitted):

(1) $W = \{ (x_1,x_2) \in X_1 \times X_2 \mid x_1 = x_1^0 \text{ or } x_2 = x_2^0 \} = X_1 \times \{ x_2^0\} \cup \{ x_1^0\} \times X_2$ with the subspace topology inherited from $X_1 \times X_2$, $w^0 = (x_1^0,x_2^0)$, $j_1(x_1) = (x_1,x_2^0)$, $j_2(x_2) = (x_1^0,x_2)$.

(2) The disjoint union of $X_1, X_2$ may be defined as $X_1 + X_2 = X_1 \times \{ 1 \} \cup X_2 \times \{ 2 \}$ which is topologized by defining $U \subset X_1 + X_2$ to be open if it has the form $U = U_1 \times \{ 1 \} \cup U_2 \times \{ 2 \}$ with open $U_i \subset X_i$. The $X_i \times \{ i \}$ are copies of $X_i$ which are made disjoint via the second factor. Note that this also applies for $X_1 = X_2$.

Now define $W$ as the quotient space $(X_1 \times \{ 1 \} \cup X_2 \times \{ 2 \})/\sim$, where $(x_1^0,1) \sim (x_2^0,2)$. Set $w^0 = [x_1^0] = [x_2^0]$, $j_i(x_i) = [(x_i,i)]$.

(3) This variant only works in the following special case: Let $X_i$ be closed subspaces of a space $X$ such that $X_1 \cap X_2$ contains a single point $x^0$. Then we may take $(X_1 \cup X_2,x^0)$ with inclusions $j_i : (X_i,x^0) \hookrightarrow (X_1 \cup X_2,x^0)$ as the a one-point union of $(X_i,x^0)$. This construction is particularly "intuitive" and can be used in Munkres' exercise. See the other answers to your question. Also see exercise 2 on p. 438 of Munkres.

Note that for the approach in (3) it is essential that both $X_i$ are closed in $X$. Consider for example $X = [0,1], X_1 = [0,1), X_2 = \{ 0,1 \}$. Then ordinary union does not yield a one-point union.

The universal property has a number of consequences (this could also be seen by using the explicit constructions, but it is nice to work only with the universal property because it indicates how to construct the desired system).

(a) Both $j_i$ are embeddings. That is, both $(\tilde{X}_i,\tilde{x}_i^0) \subset (W,w^0)$ are homeomorphic copies of $(X_i,x_i^0)$.

In fact, let $c_{21} : (X_2,x_2^0) \to (X_1,x_1^0)$ denote the map with $c_{21}(X_2) = \{ x_1^0) \}$. Then there exists a unique base point preserving $F: (W,w^0) \to (X_1,x_1^0)$ such that $F \circ j_1 = id : (X_1,x_1^0) \to (X_1,x_1^0)$ and $f \circ j_2 = c_{21}$; similarly for $j_2$. This shows moreover that $j_i$ embeds $(X_i,x_i^0)$ as a pointed retract into $(W,w^0)$.

(b) $\tilde{X}_1 \cup \tilde{X}_2 = W$.

Let $(\tilde{W},\tilde{w}^0) = (\tilde{X}_1 \cup \tilde{X}_2,w^0)$ and let $\tilde{j}_i : (X_i,x_i^0) \to (\tilde{W},\tilde{w}^0)$ be defined by $\tilde{j}_i(x_i) = j_i(x_i)$. It is easy to see that this system has again the universal property, thus the inclusion $(\tilde{W},\tilde{w}^0) \to (W,w^0)$ is a homeomorphism. This implies (b).

(c) $\tilde{X}_1 \cap \tilde{X}_2 = \{ w^0 \}$.

Define $(Z,z^0) = (\{ 0, 1 \}, 0)$ with the trivial topology on $\{ 0, 1 \}$. Let $f_1 : (X_1,x_1^0) \to (Z,z^0), f_1(x_1) = 0$ for all $x_1$ and $f_2 : (X_2,x_2^0) \to (Z,z^0)$, $f_2(x_2^0) = 0$, $f_2(x_2) = 1$ for $x_2 \ne x_2^0$. Both are are continuous maps. There exists a unique map $F : (W,w^0) \to (Z,z^0)$ such that $F \circ j_i = f_i$. For $w \in \tilde{X}_1 \cap \tilde{X}_2$ we can write $w = j_1(x_1) = j(x_2)$ with unique $x_i \in X_i$. Then $0 = f_1(x_1) = F(w) = f_2(x_2)$, hence $x_2 = x_2^0$ and $w = w^0$.