Claim: the homomorphism $$ (p_X)_* ×(p_Y)_*: π_1(X × Y,(x_0, y_0)) → π_1(X, x_0) × π_1(Y, y_0) $$ is an isomorphism.

In category theory, product of objects is defined by the universal property that for any object $Z$ and maps $f: Z → X$ and $g: Z → Y$, there is a unique map $h: Z → X × Y$ such that $p_X∘h=f$ and $p_Y∘h=g$.

Applying this to the fundamental group, we have that the product of the fundamental groups is the unique group that satisfies the universal property of the product of groups. This is the direct product of the fundamental groups, and the isomorphism follows.

A property of the product topology is that a map $f: Z → X×Y$ is continuous if and only if both $p_X◦f:Z→X$ and $p_Y◦f:Z→Y$ are continuous.

We map a loop $f$ in $X×Y$ based at $(x_0, y_0)$ to a pair of loops $g$ in $X$ and $h$ in $Y$ based at $x_0$ and $y_0$. The map is surjective: for $[l_1]∈π_1(X,x),[l_2]∈π_1(Y,y)$, let$$l_1× l_2:t↦(l_1(t),l_2(t))$$then \begin{array}l p_X(l_1× l_2)=l_1\\ p_Y(l_1× l_2)=l_2 \end{array} Similarly, a homotopy $f_t$ of a loop in $X×Y$ is equivalent to a pair of homotopies of the projection loops in $X$ and $Y$. Thus we obtain a bijection $π_1(X×Y,(x_0,y_0)) ≈ π_1(X,x_0)×π_1(Y,y_0)$. This is obviously a group homomorphism, and hence an isomorphism.

Applying this to the torus, we have an isomorphism $π_1(S^1×S^1)≈ℤ×ℤ$. Under this isomorphism a pair $(p,q)∈ℤ×ℤ$ corresponds to a loop that winds $p$ times around one $S^1$ factor of the torus and $q$ times around the other $S^1$ factor.

• Prove that there is no retraction map $r: D^2 → S^1$.
• Our aim here is to show that any map $f: D^2 → D^2$ has a fixed point. Suppose that, on the contrary, $f$ has no fixed point; in other words $f(x) ≠ x$ for all $x ∈ D^2$.
  Use the pairs $(x, f(x))$ to construct a retraction $D^2 → S^1$. Thus, we deduce that any map $D^2 → D^2$ must have a fixed point. (This fact has many applications outside of topology. For example, it can be used to show that certain differential equations always have a solution.)

• Let $f_0$ be any loop in $S^1$ based at $x_0$. In the convex region $D^2$ there is a homotopy of $f_0$ to a constant loop.
  Error! Click to view log.

  Since ${r|}_{S^1}=$id, the composition $r f_t$ is then a homotopy in $S^1$ from $r f_0=f_0$ to the constant loop at $x_0$. But this contradicts the fact that $π_1(S^1)$ is non-trivial.

  Another way: If $r:D^2→S^1$ is a retraction, $ri=$id, then $r_*i_*=$id, so $i_*$ is injective. But $i_*:ℤ→0$ cannot be injective.

• Define a map $r:D^2→S^1$ by letting $r(x)$ be the point of $S^1$ where the ray in $R^2$ starting at $f(x)$ and passing through $x$ leaves $D^2$.
  Clearly $r$ is continuous [define $r(x)=x+λ(x-f(x)),{‖r(x)‖}=1,λ≥0$].
  If $f(x) ≠ x$, $r$ is well-defined. ${r|}_{S^1}=$id. Thus $r$ is a retraction of $D^2$ onto $S^1$.
  Error! Click to view log.

Suppose $f: ℝ → ℝ^n(n>1)$ is a homeomorphism, then $f(ℝ-\{0\})=ℝ^n-\{f(0)\}$.
ℝ∖{0} is disconnected while $ℝ^n∖\{f(0)\}$ is connected, contradiction.

Suppose $f: ℝ^2 → ℝ^n(n>2)$ is a homeomorphism, $f$ induces an isomorphism between fundamental groups. For a point $a∈ℝ^n$, $ℝ^n-\{a\}$ can homotopic retract to $S^{n-1}$ [in fact it is homeomorphic to $S^{n-1}×ℝ$ via $x↦\left(\frac{x-a}{|x-a|},{|x-a|}\right)$], so Q3 implies that $π_1\left(ℝ^n-\{a\}\right)$ is isomorphic to $π_1\left(S^{n-1}\right)×π_1(ℝ)≅π_1\left(S^{n-1}\right)$.
$π_1(S^{n-1})=ℤ$ for $n=2$ and trivial for $n>2$, by III.31,32 [or from Q1 + sheet 1 Q9].
Hence $π_1\left(ℝ^n-\{a\}\right)=\cases{ℤ&for $n=2$\\1&for $n>2$}$ They are not isomorphic.

Let $g$ be a non-trivial reduced word. Take a reduced subword $h$ of $g$ of maximal length such that $g=hxh^{-1}$ for some reduced word $x$. Because $g$ is reduced, $x≠1$ and by maximality of $h$, first and last letter of $x$ are not inverse, so $xx$ is reduced, $g^n=hx^nh^{-1}$ is reduced; hence $g$ has infinite order.

Suppose $h∈F(S)∖\{e\}$. Write $h$ as reduced word $h=g_1…g_n$ where each $g_i∈S∪S^{-1}$. Take $g∈S∪S^{-1}$ such that $g≠g_n^{-1}$. Then$$gg_1^{-1}h=gg_2…g_n$$ has reduced length less than $n+2$. On the other hand,$$hgg_1^{-1}=g_1…g_ngg_1^{-1}$$is a reduced word of length $n+2$, so $gg_1^{-1}h≠hgg_1^{-1}$, so $h∉Z(F(S))$.

• Let $F$ be the free group on the three generators $x, y$, and $z$. For non-zero integers $r$, $s$, and $t$, show that the subgroup of $F$ generated by $x^r, y^s$, and $z^t$ is freely generated by these elements.
• Let $H$ be the subgroup of $F(\{x, y\})$ generated by $x^2, y^2, x y$, and $y x$. Show that $H$ is not freely generated by these elements.

• By universal property of free group, id: $\{x^r,y^s,z^t\}→⟨x^r,y^s,z^t⟩$ induces a group homomorphism $ϕ:F(\{x^r,y^s,z^t\})→⟨x^r,y^s,z^t⟩$ which is an isomorphism.
• $x^2(yx)^{-1}y^2(xy)^{-1}=1$ is a non-trivial relation of generators, so $H$ is not free.

$H$ is the kernel of the homomorphism $F(\{x,y\})→ℤ/2ℤ,x↦1,y↦1$. The corresponding covering space of $S^1∨S^1$ is Error! Click to view log.