Definition 2.1 Let $X$ be a space and let $x_0 \in X$ be an element in that space. We call a path in $X$ that begins and ends at $x_0$ a loop based at $\mathbf{x_0}$. The set of path homotopy classes of loops based at $x_0$, with operation $\ast$ is called the fundamental group or first homotopy group of $X$ relative to the base point $x_0$.



We denote this group with the notation $\pi_1 (X,x_0$)



We can easily verify that the fundamental group is indeed a group:



     Closure: Let $f$ and $g$ be two loops based at $x_0$. Their concatenation will also be a loop based at $x_0$.



     Associativity: We proved that the operation $*$ is associative in the l\ast section.



     Identity: If $f$ is a loop based at $x_0$, its inverse will be $\tilde{f}$ will be the loop traveling the path of $f$ in the opposite direction.



     Inverse: the trivial loop $e_{x_0}$ which is equal to $x_0$ for all $t$ is the identity element for this group.



Definition 2.2 Let $\alpha$ be a path in $X$ from $x_0$ to $x_1$. We define the map $$\hat{\alpha} : \pi_1(X,x_0) \longrightarrow \pi_1(X,x_1)$$ by the equation $$\hat{\alpha} \big([f] \big) = [\tilde{\alpha}] \ast [f] \ast [\alpha]$$



Below is an example of what the resulting loop may look like. $a$ and $\tilde{a}$ are drawn separated for visual clarity. In reality they would follow the same curve







Theorem 2.3 The map $\hat{\alpha}$ is a group isomorphism.



Proof: First, we will show that $\hat{\alpha}$ is a homomorphism. By the previous definition, we have $$\hat{\alpha}[f] \ast \hat{\alpha} [g] = \big( [\tilde{\alpha}] \ast [f] \ast [\alpha] \big) \ast \big( [\tilde{\alpha}] \ast [g] \ast [\alpha] \big)$$ $$= [\tilde{\alpha}] \ast [f] \ast [g] \ast [\alpha] = \hat{\alpha} \big( [f] \ast [g]\big)$$ Therefore, the map $\hat{\alpha}$ is an homomorphism.



Next, we need to prove that $\hat{\alpha}$ is bijective. To show this, we will will show that there exists an inverse function of $\hat{\alpha}$ given by $$\hat{\beta} : \pi_1(X,x_1) \longrightarrow \pi_1(X,x_2)$$ such that for all $[f] \in \pi_1(X,x_0)$ we have $$\hat{\beta} \big( \hat{\alpha} [f] \big) = [f] $$ and for all $[h] \in \pi_1(X,x_1)$ $$\hat{\alpha} \big( \hat{\beta}[h]\big) = [h]$$ To find this inverse, let $\beta = \tilde{\alpha}$ be a path from $x_1$ to $x_0$. Then, $$\hat{\beta} \big( [h] \big) = [\tilde{\beta}] \ast [h] \ast [\beta] = [\alpha] \ast [h] \ast [\tilde{\alpha}]$$ is the inverse of $\hat{\alpha}$



Thus for all $[h] \in \pi_1(X,x_1)$ we have $$\hat{\alpha} \big( \hat{\beta}[h]\big) = \hat{\alpha} \big([\tilde{\beta}] \ast [h] \ast [\beta] \big) $$ $$=\hat{\alpha}\big( [\alpha] \ast [h] \ast [\tilde{\alpha}] \big)$$ $$= [\tilde{\alpha}] \ast [\alpha] \ast [h] \ast [\tilde{\alpha}] \ast [\alpha] $$$$= [h] $$ By the same argument we have that for all $[f] \in \pi_1 (X,x_0)$ $$\hat{\beta} \big( \hat{\alpha} [f] \big) = [f] $$ Therefore, the map $\hat{\alpha}$ is a group isomorphism.



Corollary 2.4 If X is path connected and $x_0,x_1 \in X$, then $$\pi_1 (X,x_0) \simeq \pi_1(X,x_1)$$







Definition 2.5 A space $X$ is said to be simply connected if it the following two conditions hold:



     I. $X$ is path connected



     II. Every loop based at $x_0$ can be continuously transformed into a single point.



In other words, condition II states every loop must be homotopic to the constant loop $e_{x_0}$. We denote this as $$\pi_1(X,x_0) = 0$$ or by referring to $\pi_1(X,x_0)$ as the trivial group.











Lemma 2.6 Let X be a simply connected space. Any two paths in X which share the same initial and final points are path homotopic.



Proof: Let $\alpha$ and $\gamma$ be two paths from $x_0$ to $x_1$. Then, $\alpha * \tilde{\gamma}$ is defined and is a loop on X based at $x_0$.



Since X is simply connected, this loop is path homotopic to the constant loop at $x_0$. This means that $$[\alpha \ast \tilde{\gamma}] = [e_{x_0}] $$ which implies $$[\alpha] = [\gamma]$$



Definition 2.7 Let $$h: (X,x_0) \longrightarrow (Y,y_0)$$ be a continuous map that maps space $X$ to space $Y$, and specifically the point $x_0 \in X$ to $y_0 \in Y$.



We define the function $$h_{\ast}: \pi_1(X,x_0) \longrightarrow \pi_1(Y,y_0)$$ by the equation $$h_{\ast}\big([f] \big) = [h \circ f]$$ where $[f] \in \pi_1(X,x_0) $.



Below is an example of what $h_{\ast}\big([f] \big)$ may look like.











Theorem 2.8 $h_{\ast}$ is a homomorphism.



Proof: First we must verify that $h_{\ast}$ is well defined on path homotopy classes. In other words, if we swap $f$ with another loop $f'$ in the same homotopy class, we should get that $h \circ f \simeq_p h \circ f' $. This will guarantee that we get the same answer no matter what representative of $[f]$ we choose. This isn't too difficult to demonstrate:



Let $F$ be a path homotopy between the paths $f$ and $f'$. Then, $h \circ F$ is a path homotopy between $h \circ f$ and $h \circ f'$, giving us $[h \circ f] = ]h \circ f']$.



Now, to show that $h_{\ast}$ is a homomorphism, consider the equation: $$h_{\ast} \big([f] \ast [g] \big) = h_{\ast} \big([f \ast g] \big)$$ $$ = [h \circ (f \ast g)] = [(h \circ f) \ast (h \circ g)] $$ $$= [h \circ f] \ast [h \circ g] = h_{\ast} \big([f]\big) \ast h_{\ast} \big( [g] \big)$$ Therefore, $h_*$ is a homomorphism. Now it makes sense to give $h_*$ its proper name:



Definition 2.7.2. The map $h_{\ast}$ is called the homomorphism induced by $\boldsymbol{h}$, relative to the base point $x_0$.



Now, to prove some properties of $h_*$.



Theorem 2.9 This theorem has two parts.

     a. If $$h: (X,x_0) \longrightarrow (Y,y_0)$$ and $$k: (X,x_0) \longrightarrow (Y,y_0)$$ are both continuous maps, then $$(k \circ h)_{\ast} = k_{\ast} \circ h_{\ast}$$      b. If $$i: (X,x_0) \longrightarrow (X,x_0)$$ is the identity map, then $$i_{\ast}: \pi_1(X,x_0) \longrightarrow \pi_1(X,x_0)$$ is the identity homomorphism.



Proof: Both parts follow directly from the definition of $h_*$.      a. Given $k$ and $h$ as defined above, we have $$\big(k \circ h \big)_{\ast} \big( [f] \big) = [ (k \circ h) \circ f] = [ k \circ h \circ f]$$ and $$\big(k_{\ast} \circ h_{\ast} \big) \big( [f] \big) = k_{\ast}\big( [h \circ f] \big) = [k \circ (h \circ f) ] = [ k \circ h \circ f]$$ Therefore, $(k \circ h)_{\ast} = k_{\ast} \circ h_{\ast}$.



     b. If we let $i$ be the identity map then $$i_{\ast} \big( [f] \big) = [i \circ f] = [f] $$



Corollary 2.10 If $$h:(X,x_0) \longrightarrow (Y,y_0)$$ is a homomorphism of X and Y, then $h_{\ast}$ is an isomorphism of $\pi_1 (X,x_0)$ and $\pi_1(Y,y_0)$.



Proof: We already know that $h_*$ is a homomorphism, so we just need to show that it is bijective.



Let $$k: (Y,y_0) \longrightarrow (X,x_0)$$ be the inverse of $h$. Then, $$k_{\ast} \circ h_{\ast} = (k \circ h)_{\ast} = i_{\ast} $$ where $i$ the identity map $$i: (X,x_0) \longrightarrow (X,x_0)$$ and $$h_{\ast} \circ k_{\ast} = (h \circ k)_{\ast} = j_{\ast} $$ where $j$ the identity map $$j: (Y,y_0) \longrightarrow (Y,y_0)$$ We know from the previous theorem that $i_{\ast}$ and $j_*$ are the identity homomorphisms of $\pi_1 (X,x_0)$ and $\pi_1(Y,y_0)$ respectively. Therefore, $k_*$ is the inverse of $h_*$, and $h_*$ is an isomorphism.



Problem 2.1 A subset $A \subseteq \mathbb{R}^n$ is said to be star convex if, for some point $a_0 \in A$, all line segments joining $a_0$ to other points in $A$ lie entirely in $A$.



     a. Find a star convex set that is not convex



     b. Show that if $A$ is star convex, it is simply connected.



Proof:      a. To find a star convex set that is not convex, look no further than the namesake:







     b. To show that $A$ is simply connected, we first must show it is path connected. Let $x_0, x_1 \in A$ and define the line segments from $a_0$ to $x_0$ and $x_1$ by $$f(t) = (1-t) x_0 + t a_0, \quad \quad g(t) = (1-t) a_0 + t x_1$$ where $0 \leq t \leq 1$. Since we have $$f(0) = x_0, \quad f(1) = g(0) = a_0, \quad g(1) = x_1 $$ the product $h = f* g$ will be a path from $x_0$ to $x_1$ contained entirely within $A$.



Now we need to show that $\pi_1(A,a_0)$ is the trivial group. Let $\gamma$ be a loop based at $a_0$ in $A$. We can define the straight line path homotopy $$F(s,t) = \gamma(s) (1-t) + t a_0$$ which, by the definition of star convex, must be continuous in $A$. Thus, $\gamma \simeq_p a_0$, meaning that $\pi_1(A,a_0)$ is the trivial group.



Therefore, $A$ is simply connected.



Problem 2.2 Let $\alpha$ be a path in $X$ from $x_0$ to $x_1$ and let $\beta$ be a path in $X$ from $x_1$ to $x_2$. Show that if $$\gamma = \alpha * \beta$$ then $$\hat{\gamma} = \hat{\beta} \circ \hat{\alpha}$$



Proof: By Definition ref{hat}, $$\hat{\gamma} \big( [f] ) = \left[\tilde{\gamma} \right] * \left[f \right] * \left[\gamma\right]$$ $$=\left[\widetilde{\alpha * \beta} \right] * \left[f \right] * \left[\alpha * \beta \right]$$ $$= [\tilde{\beta}] * [\tilde{\alpha}] * [f] * [\alpha] * [\beta]$$ $$= [\tilde{\beta}] * \hat{\alpha} [f] * [\beta]$$ $$= (\hat{\beta} \circ \hat{\alpha}) [f]$$ Therefore $$\hat{\gamma} = \hat{\beta} \circ \hat{\alpha}$$



Problem 2.3 Let $x_0$ and $x_1$ be points in path connected space $X$, Show that $\pi_1(X,x_0)$ is abelian if and only if for every pair of paths $\alpha$ and $\beta$ from $x_0$ to $x_1$, $\hat{\alpha} = \hat{\beta}$.



Proof      $(\Rightarrow)$ Let $\pi_1(X,x_0)$ be an abelian group. Remember that a group is abelian if its operation is commutative. In our case, if $[f]$ and $[g]$ are homotopy classes of loops based at $x_0$, we have that $$[f] * [g] = [g] * [f]$$ Now consider the two homotopies $$\hat{\alpha} \big([f] \big) = [\tilde{\alpha}] \ast [f] \ast [\alpha]$$ and $$\hat{\beta} \big([f] \big) = [\tilde{\beta}] \ast [f] \ast [\beta]$$ and take $$\left( \hat{\tilde{\beta}} \circ \hat{\alpha} \right)[f]$$ If $\hat{\alpha} = \hat{\beta}$, then $\hat{\tilde{\beta}}$ and $\hat{\alpha}$ should be inverses and this expression should be equal $[f]$. We can show that this is true by expanding the expression and then using commutativity. $$ \left( \hat{\tilde{\beta}} \circ \hat{\alpha} \right)[f] = \hat{\tilde{\beta}} \big( [\tilde{\alpha}] \ast [f] \ast [\alpha] \big)$$ $$[\beta] * [\tilde{\alpha}] * [f] * [\alpha] * [\tilde{\beta}]$$ Since $\pi_1(X,x_0)$ is abelian and $[f]$, $([\beta] * [\tilde{\alpha}])$, and $([\alpha] * [\tilde{\beta}])$ are all elements in $\pi_1(X,x_0)$, this is equal to $$[f] * [\beta] * [\tilde{\alpha}] * [\alpha] * [\tilde{\beta}]$$ $$= [f] * [\beta] * \left[e_{x_0} \right] * [\tilde{\beta}]$$ $$=[f] * [\beta] * [\tilde{\beta}]$$ $$ = [f] * \left[e_{x_0} \right]$$ $$=[f]$$ Therefore, $\hat{\alpha} = \hat{\beta}$.



     $(\Leftarrow)$ Now, assume that for every pair of paths $\alpha$ and $\beta$ from $x_0$ to $x_1$, $\hat{\alpha} = \hat{\beta}$, and let $f$ and $g$ be two loops based at $x_0$. We will choose $\alpha$ and $\beta$ in a way that wll help us prove commutativity.



Let $\gamma$ be any path from $x_0$ to $x_1$ and let $\alpha$ and $\beta$ be the paths $$\alpha = f * \gamma, \quad \quad \beta = g * \gamma$$ These are both paths from $x_0$ to $x_1$ and therefore $$\hat{\alpha} = \hat{\beta}$$ $$\Rightarrow [\tilde{\alpha}] \ast [f] \ast [\alpha] = [\tilde{\beta}] \ast [f] \ast [\beta]$$ $$\Rightarrow [\tilde{\gamma}] * [\tilde{f}] \ast [f] \ast [f] * [\gamma] = [\tilde{\gamma}] * [\tilde{g}] \ast [f] \ast [g] * [\gamma]$$ Since $[\tilde{f}] \ast [f] \ast [f] = [e_{x_0}] \ast [f] = f$, this implies that $$[f] = [\tilde{g}] \ast [f] \ast [g]$$ $$\Rightarrow [g] * [f] = [f] * [g]$$ Therefore, $\pi_1(X,x_0)$ is abelian.







Problem 2.4 Let $A \subset X$ and let $$r: X \longrightarrow A$$ be a continuous map such that $$r(a) =a \quad \quad \forall a in A$$ If $a_0 \in A$, show that $$r_*: \pi_1(X,a_0) \longrightarrow \pi_1(A,a_0)$$ is surjective.



Proof: To show that $r_*$ is surjective, we need to show that every $[g] \in \pi_1(A,a_0)$, there exists a $[f] \in \pi_1(X,a_0)$ such that $r_*([f]) = [g]$.



Now it should make sense why this would be true. Since $A \subset X$, $\pi_1(A,a_0) \subset \pi_1(X,a_0)$, restricting the domain of $r_*$ to $\pi_1(A,a_0)$ will take every element and map it to itself.



However, to be a bit more rigorous we will show that we can define an inverse function that we use as our $[f]$ which will guarantee that every $[g] \in \pi_1(A,a_0)$ is accounted for.



Let $$j:A \longrightarrow X$$ be the inclusion map $$j(a) = a \quad \quad \forall a \in A$$ with induced homomorphism $$j_*: \pi_1(A,a_0) \longrightarrow \pi_1(X,a_0)$$ Now, to show that $r$ and $j$, and by extension $r_*$ and $j_*$ are indeed inverses of each other, observe that for $a \in A$, we have $$(r \circ j) (a) = r\big(j(a) \big) = r(a) = a$$ $$\Rightarrow r \circ j = i$$ where $i$ is the identity map. Therefore, by theorem ref{identity}, we have that $$(r \circ j)_* = r_* \circ j_* = i_*$$ where $i_*$ is the identity homomorphism $$i_{\ast}: \pi_1(A,a_0) \longrightarrow \pi_1(A,a_0)$$ So if we let $[f] = j_*[(g])$, we have $$r_* ([f]) = r_*\big(j_*([g])\big) = i_*([g]) = [g]$$ Therefore, $r_*$ is surjective.



Problem 2.5 A trivial homomorphism is a homomorphism $$\phi: X \longrightarrow Y$$ between two groups that maps every element in $X$ to the identity element in $Y$.



Let $A$ be a subspace of $\mathbb{R}^n$, and let $$h: (A,a_0) \longrightarrow (Y, y_0)$$ Show that if $h$ is extendable to a continuous map of $\mathbb{R}^n$ into $Y$, then $h_*$ is the trivial homomorphism.



Proof: Let $H$ be an extension of $h$ on $\mathbb{R}^n$. The homomorphism induced by $H$ is given by $$H_*: \pi_1(\mathbb{R}^n,a_0) \longrightarrow \pi_1(Y, y_0)$$ Since $\mathbb{R}^n$ is simply connected, $\pi_1(\mathbb{R}^n,a_0) = 0$. This means that the domain of $H_*$ is only $[x_0]$, which by definition gets mapped to $[y_0]$.



Now, since $h$ is just $H$ restricted to $A$, $h_*$ must also map only to $[y_0]$, which is the identity element of $\pi_1(Y, y_0)$.



To be more rigorous, if we define $j: A \longrightarrow \mathbb{R}^n$ to be inclusion map as defined in Problem 4, we get the following diagram.







Therefore we have that $$h(a) = (H \circ j)(a)$$ $$\Rightarrow h_* = H_* \circ j_*$$ $$h_*\Big( \pi_1(A,a_0) \Big) = (H_* \circ j_*) \Big( \pi_1(A,a_0) \Big) $$ $$= H_* \Big( \pi_1(\mathbb{R}^n,a_0)\Big) = [y_0]$$ Therefore, $h_*$ is a trivial homomorphism.



Problem 2.6 Show that if $X$ is path connected, the homomorphism induced by a continuous map is independent of base point up to isomorphisms of the group involved.



More precisely, let $$h: X \longrightarrow Y$$ be continuous with $$h(x_0) = y_0, \quad \quad h(x_1) = y_1$$ and define two homomorphisms with different base points, both induced by $h$ $$h_{x_0}: \pi_1(X,x_0) \longrightarrow \pi_1(Y,y_0)$$ and $$h_{x_1}: \pi_1(X,x_1) \longrightarrow \pi_1(Y,y_1)$$ Let $\alpha$ be a path from $x_0$ to $x_1$ and let $\beta = h \circ \alpha$.



Show that $$\hat{\beta} \circ (h_{x_0})_* = (h_{x_1})_* \circ \hat{\alpha}$$ In other words, show that the diagram below commutes



>

and therefore $h_{x_0}$ and $h_{x_1}$ must be isomorphic.



Proof: Despite the lengthy setup, this proof comes directly from applying definitions ref{hat} and ref{star. By proving $$\hat{\beta} \circ (h_{x_0})_* = (h_{x_1})_* \circ \hat{\alpha}$$ we will show that this is the required isomorphism between $h_{x_0}$ and $h_{x_1}$.



Let $f \in \pi_1(X,x_0)$ and observe that: $$\Big[(h_{x_1})_* \circ \hat{\alpha}\Big]\big([f] \big) $$$$= (h_{x_1})_* \Big([\tilde{\alpha}] * [f] * [\alpha] \Big)$$ $$= [h \circ \tilde{\alpha}] * [h \circ f] * [h \circ \alpha]$$ $$= [\tilde{\beta}] * [h \circ f] * [\beta]$$ $$= \hat{\beta} \circ \Big([h \circ f] \Big)$$ $$= \hat{\beta} \circ \Big(\left(h_{x_0}\right)_* \big([f] \big) \Big)$$ $$= \left[\hat{\beta} \circ (h_{x_0})_*\right] \big([f] \big)$$ Below are two graphs that visualize $\Big[(h_{x_1})_* \circ , \hat{\alpha}\Big]\big([f] \big)$ and $\left[\hat{\beta} \circ (h_{x_0})_*\right] \big([f] \big)$ respectively.



Note that both start at $x_0$ and end at $y_1$





Hence, $h_{x_0}$ and $h_{x_1}$ are equivalent up to isomorphism.



Problem 2.7 Let $A$ be a topological group with operation $\cdot$ and identity element $x_0$. Let $\Omega (A,x_0)$ denote the set of all loops in $A$ based at $x_0$.



Let $f,g \in \Omega (A,x_0)$ and define the loop $(f \otimes g) (s)$ by $$(f \otimes g) (s) = f(s) \cdot g(s)$$ where $0 \leq s \leq 1$. Show that:



     a. $\Omega (A,x_0)$ is a group with operation $\otimes$



     b. $\otimes$ induces a group operation on $\pi_1 (A,x_0)$



     c. the group operations $*$ and $\otimes$ on $\pi_1 (A,x_0)$ are the same



     d. $\pi_1(A, x_0)$ is abelian



Proof:



     a. For this proof, we will make use of fact that $\cdot$ is the group operation for $A$ and therefore must satisfy each of the following properties for all elements in $A$.



     Closure: $(f \otimes g)(s) = f(s) \cdot g(s)$ is a loop by definition, which is contained entirely in $A$ since $f(s), g(s) in A$.



We also know that $f(0) = g(0) = f(1) = g(1) = x_0$, which is the identity element of $A$. Therefore we have that $$(f \otimes g)(0) = f(0) \cdot g(0) = x_0 \cdot x_0 = x_0$$ and $$(f \otimes g) (1) = f(1) \cdot g(1) = x_0 \cdot x_0 = x_0$$ Thus, $(f \otimes g) (s) \in \Omega (A,x_0)$.



     Associativity: Letting $f,g, h \in \Omega(A,x_0)$, we have: $$\Big[f \otimes (h \otimes g) \Big](s) = f(s) \cdot (h \otimes g) (s)$$ $$= f(s) \cdot \Big(h(s) \cdot g(s) \Big)= \Big(f(s) \cdot h(s) \Big) \cdot g(s)$$ $$ = \Big[(f \otimes h) \otimes g \Big](s)$$      Identity: Define the trivial loop $$e_{x_0}(s) = x_0 \quad \quad \forall s \in [0,1]$$ We have that $e_{x_0} \in \Omega(A,x_0)$ and we can demonstrate that it serves as the identity element: $$\left(f \otimes e_{x_0} \right)(s) = f(s) \cdot e_{x_0}(s) = f(s) \cdot x_0 = f(s)$$ $$\left( e_{x_0} \otimes f\right)(s) = e_{x_0}(s) \cdot f(s)= x_0 \cdot f(s)= f(s)$$      Inverse: Let $f \in \Omega(A,x_0)$ and consider $f^{-1}$ defined by $f^{-1}(s) = \big( f(s) \big)^{-1}$, or the inverse of $f(s)$ in A.



Then, $f^{-1}$ is continuous and $$f^{-1}(0) = f^{-1}(1) = \big( f(0) \big)^{-1} = \big( f(1) \big)^{-1} = x_0^{-1} = x_0$$ Now, applying the operation $\otimes$, $$\left( f \otimes f^{-1}\right)(s) = f(s) \cdot f^{-1}(s) = f(s) \cdot \big( f(s) \big)^{-1} = x_0$$ $$\Rightarrow f \otimes f^{-1} = e_{x_0}$$ $\left( f^{-1} \otimes f\right)(s) $ will give the same result, verifying that $f^{-1}$ is the inverse of $f$.



     b. Define the group operation $$[f] \otimes [g] = [f \otimes g]$$ We must show that this operation is well-defined when applied to homotopy classes. In other words, if $f \simeq f'$ and $g \simeq g'$, then it must be true that $$f \otimes g \simeq_p f' \otimes g' $$ First, we will define the path homotopies $F$ between $f$ and $f'$, and $G$ between $g$ and $g'$. These are given by $$F: I \times I \longrightarrow A $$ such that $$F(s,0) = f(s), \quad \quad F(s,1) = f'(s) $$ $$F(0,t) = F(1,t) = x_0$$ and $$G: I \times I \longrightarrow A $$ such that $$G(s,0) = g(s), \quad \quad G(s,1) = g'(s) $$ $$G(0,t) = G(1,t) = x_0$$ Now we can define a path homotopy between $f \otimes g$ and $ f' \otimes g' $. Define $$H: I \times I \longrightarrow A$$ by $$H(s,t) = F(s,t) \cdot G(s,t)$$ Now we have $$H(s,0) = F(s,0) \cdot G(s,0) = f(s) \cdot g(s) = (f \otimes g) (s)$$ $$H(s,1) = F(s,1) \cdot G(s,1) = f(s) \cdot g(s) = (f' \otimes g') (s)$$ and $$H(0,t) = F(0,t) \cdot G(0,t) = x_0 \cdot x_0 = x_0$$ $$H(1,t) = F(1,t) \cdot G(1,t) = x_0 \cdot x_0 = x_0$$ which verifies that $H$ is a path homotopy between $(f \otimes g)$ and $(f' \otimes g')$.



Therefore $f \otimes g \simeq_p f' \otimes g'$ and $$[f] \otimes [g] =[f \otimes g] = [f' \otimes g'] = [f'] \otimes [g']$$ is well-defined.







     c. Using the trivial loop $e_{x_0}$, we can rewrite $f \otimes g$ as $$(f \otimes g)(s) = \Big((f * e_{x_0}) \otimes (e_{x_0} * g) \Big) (s) $$ Looking more closely at each term ,we see that $$\big(f * e_{x_0})(s) = \begin{cases} f(2s) & s \in \left[ 0,\frac{1}{2}\right] \\ e_{x_0}(2s-1) & s \in \left[\frac{1}{2}, 1\right] \end{cases} $$ $$ = \begin{cases} f(2s) & s \in \left[ 0,\frac{1}{2}\right]\\ x_0 & s \in \left[\frac{1}{2}, 1\right] \end{cases} $$ and $$\big( e_{x_0}*g)(s) = \begin{cases} e_{x_0}(2s) & s \in \left[ 0,\frac{1}{2}\right]\\ g(2s-1) & s \in \left[\frac{1}{2}, 1\right] \end{cases} $$ $$ = \begin{cases} x_0 & s \in \left[ 0,\frac{1}{2}\right]\\ g(2s-1) & s \in \left[\frac{1}{2}, 1\right] \end{cases} $$ Combining the two, we get: $$(f \otimes g)(s) = \Big((f * e_{x_0}) \otimes (e_{x_0} * g) \Big) (s) = \begin{cases} f(2s) \cdot x_0 & s \in \left[ 0,\frac{1}{2}\right]\\ x_0 \cdot g(2s-1) & s \in \left[\frac{1}{2}, 1\right] \end{cases} $$ $$ = \begin{cases} f(2s) & s \in \left[ 0,\frac{1}{2}\right]\\ g(2s-1) & s \in \left[\frac{1}{2}, 1\right] \end{cases} \quad = (f*g)(s)$$ Therefore $$f\otimes g = f * g$$



     d. Using our result from part (c), we have $$(f\otimes g)(s) = \Big((f * e_{x_0}) \otimes (e_{x_0} * g) \Big)(s)$$ $$=\begin{cases} f(2s) \cdot x_0 & s \in \left[ 0,\frac{1}{2}\right]\\ x_0 \cdot g(2s-1) & s \in \left[\frac{1}{2}, 1\right] \end{cases} $$ Since $x_0$ is the identity element, this is equal to $$=\begin{cases} x_0 \cdot f(2s) & s \in \left[ 0,\frac{1}{2}\right]\\ g(2s-1) \cdot x_0 & s \in \left[\frac{1}{2}, 1\right] \end{cases} $$ What happens if we change the order of terms like this? Now we have $$\big(e_{x_0} * f)(s) = \begin{cases} x_0 & s \in \left[ 0,\frac{1}{2}\right]\\ f(2s-1) & s \in \left[\frac{1}{2}, 1\right] \end{cases} $$ and $$\big(g* e_{x_0})(s) = \begin{cases} g(2s) & s \in \left[ 0,\frac{1}{2}\right]\\ x_0 & s \in \left[\frac{1}{2}, 1\right] \end{cases} $$ So $$\begin{cases} x_0 \cdot f(2s) & s \in \left[ 0,\frac{1}{2}\right]\\ g(2s-1) \cdot x_0 & s \in \left[\frac{1}{2}, 1\right] \end{cases} $$ $$= \Big((e_{x_0} * g) \otimes (f * e_{x_0}) \Big)(s)$$ $$= (g \otimes f)(s)$$ Therefore, $f \otimes g = g\otimes f$, and $\pi_1(A,x_0)$ with operation $\otimes$ is abelian.