Applications to surfaces

Let $p_1 : X \times Y \rightarrow X$, $p_2 : X \times Y \rightarrow Y$ be the projections, and define $P : \pi_1(X \times Y, (x,y)) \rightarrow \pi_1(X,x) \oplus \pi_1(Y,y)$ by $P(\alpha ) = ((p_1)_*(\alpha ), (p_2)_*(\alpha ))$. Define $Q : \pi_1(X,x) \oplus \pi_1(Y,y) \rightarrow \pi_1(X \times Y, (x,y))$ by $Q([f],[g]) = [h]$, where $h(t) = (f(t), g(t))$. Show that $Q$ is a homomorphism and is an inverse for $P$.

Show that if $g_0, g_1 : [a,b] \rightarrow \mathbb {R}^{2}$ are maps with $g_0(a) = g_1(a)$, $g_0(b) = g_1(b)$, then $g_0$ is homotopic to $g_1$ relative to the end points.

Using the notation of the previous proof (the computation of $\pi_1(S^2, 1)$, writing $S^2 = A \cup B$ with $A = S^2 \setminus \left\{ 1\right\}$, $B = S^2 \setminus \left\{ -1\right\}$, where a loop $f$ at $1$ is subdivided by a Lebesgue number argument so that each subinterval maps into $A$ or $B$), show that if $g : [a,b] \rightarrow A$ has $g(a), g(b) \neq -1$, then $g$ is homotopic to $g'$ relative to the end points, where $g'([a,b]) \subset B$.

Show that if $g \vert [a,b]$ is homotopic to $g' \vert [a,b]$ relative to the end points (where $g : [0,1] \rightarrow X$ is a loop at $x$), then $g$ is homotopic (as loops) to $g''$ where $g'' \vert [a,b] = g' \vert [a,b]$ and $g'' \vert [0,1] \setminus (a,b) = g \vert [0,1] \setminus (a,b)$.

Prove Theorem 3.4.2.

Theorem 3.4.2: Suppose that $A$, $B$, and $A \cap B$ are path connected open sets in $X = A \cup B$ and $c \in A \cap B$. Then if $\pi_1(A,c)$ and $\pi_1(B,c)$ are trivial groups, then $\pi_1(X,c)$ is also the trivial group.

Fill in the details to show that $\pi_1(P, [1]) \simeq \mathbb {Z}^{2}$, where $P$ denotes the projective plane, using the following outline.

Use the description of $\mathbb {Z}^{2}$ as $\left\{ -1,1\right\}$ with the usual multiplication in $\mathbb {R}$. Let $p : S^2 \rightarrow P$ be the quotient map, where $x \sim -x$. Using compactness of $P$, take a cover $U_1, \ldots , U_n$ of $P$ by path-connected open sets with $p^{-1}(U_i) = U_{i1} \cup U_{i2}$, where $p \vert U_{ij} : U_{ij} \rightarrow U_i$ is a homeomorphism and $T(U_{i1}) = U_{i2}$ for the antipodal map $T(x) = -x$. Given a loop $f$ at $[1]$, use a Lebesgue number argument (as in the computation of $\pi_1(S^1,1)$) to show $f$ has a unique lifting $\tilde{f} : I \rightarrow S^2$ with $\tilde{f}(0) = 1$. Define $\bar{h} : \pi_1(P,[1]) \rightarrow \mathbb {Z}^{2} = \left\{ -1,1\right\}$ by

Show that $\bar{h}$ is well defined (using a lifted homotopy $\tilde{F}$ and the unique path-lifting property), that $\bar{h}$ is a homomorphism (lifting $f * g$ by $\tilde{f} * (T\tilde{g})$ when $\tilde{f}(1) = -1$, using that $pT = p$), that $\bar{h}$ is onto (by explicitly constructing a path in $S^2$ from $1$ to $-1$), and that $\bar{h}$ is $1$–$1$ (using $\pi_1(S^2, 1) \simeq \left\{ e\right\}$ and an argument analogous to the one used for $\pi_1(S^1,1)$).

Verify that $(\mathbb {R}, p, S^1)$, $(S^2, p, P)$ are covering spaces, where $p : \mathbb {R} \rightarrow S^1$, $p(x) = (\cos 2\pi x, \sin 2\pi x)$, and $p : S^2 \rightarrow P$ is the quotient map identifying antipodal points.

Verify that $T/(z,w) \sim h(z,w)$ is homeomorphic to $K$ where we use our original definition of $K$ as a quotient of $D^1 \times D^1$.

Here $T = S^1 \times S^1$, and $h : T \rightarrow T$ is the homeomorphism given by $h(z,w) = (-z, \bar{w})$, where we regard $z, w$ as complex numbers and $\bar{w}$ is the complex conjugate of $w$.

Give a formula for the map $(D^2, \frac{1}{2}D^2) \rightarrow (D^2, 0)$ sending $S^1$ to $S^1$ via the identity and the circle of radius $\frac{1}{2}$ to $0$, which sends radial lines to radial lines.

In linear algebra, it is shown that there is no surjective vector space homomorphism from $\mathbb {R}^{2}$ onto $\mathbb {R}^{4}$ (since the dimension of the image space of a homomorphism is always less than or equal to the dimension of the domain space). Use this to show that there is no group homomorphism from $\mathbb {Z} \oplus \mathbb {Z}$ onto $\mathbb {Z} \oplus \mathbb {Z} \oplus \mathbb {Z} \oplus \mathbb {Z}$.