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.For the surjectivity, let $([α],[β]) ∈ π_1(X, x_0) × π_1(Y, y_0)$. Then $[(α, β)] ∈ π_1(X ×Y,(x_0, y_0))$, and since $α=p_X∘(α, β)$ and $β=p_Y∘(α, β)$, we get that $(p_X)_*[(α, β)]=[α]$ and $(p_Y)_*[(α, β)]=[β]$. Therefore, $([α],[β])$ is in the image of $(p_X)_* ×(p_Y)_*$ and the map is surjective.
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.
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.
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 $h∉Z(F(S))$.
$H$ 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