$\Theta \rho ϵ\eta \Pi \alpha \tau \pi$

Out of the gate, first be sure to note that $F\circ \alpha$ for any path $\alpha$ is still a path, as it remains continuous as the composition of two continuous functions and still maps out of $[0,1]$. Now we start by supposing that ${\gamma }_a\sim {\gamma }_b$, ie there exists some homotopy $H$ between them going from ${\gamma }_a$ to ${\gamma }_b$.

Now to show that $F\circ {\gamma }_a$ and $F\circ {\gamma }_b$ are homotopic, we just have to construct a function $G:I\times I\to Y$ with the required properties of a homotopy, but if we simply do $G(s,t)=F\left(H(s,t)\right)$ which is continuous as a composition of continuous functions, then we can verify that: $$G(s,0)=F\left(H(s,0)\right)=F\left({\gamma }_a\left(s\right)\right)=F\circ {\gamma }_a\left(s\right)$$ and similarly $$G(s,1)=F\left(H(s,1)\right)=F\left({\gamma }_b\left(s\right)\right)=F\circ {\gamma }_b\left(s\right)$$ also $$G(0,t)=F\left(H(0,t)\right)=F\left({\gamma }_a\left(0\right)\right)=F\circ {\gamma }_a\left(0\right)$$ and finally that $$G(1,t)=F\left(H(1,t)\right)=F\left({\gamma }_a\left(1\right)\right)=F\circ {\gamma }_a\left(1\right)$$ as needed.

Suppose that $A,B,C:X\to Y$ are continuous, first we prove reflexivity, so we show that $A\sim A$, to do so we must construct a continuous function between them as specified here, but we can take $H(x,t)=A\left(x\right)$ which satisfies the requirements and is also continuous as its equal to $A$.

Now suppose that $A\sim B$ and let's prove that $B\sim A$, since $A\sim B$ then we know that there exists $H:X\times [0,1]\to Y$ such that $H(x,0)=A\left(x\right)$ and $H(x,1)=B\left(x\right)$, then all we have to is consider the function $H^{\prime }(x,t)=H(x,1-t)$ which is still continuous with the desired properties.

Finally suppose that $A\sim B$ and that $B\sim C$ we must prove $A\sim C$ to show that transitivity still holds true, so now we have $H_1,H_2$ between the previous two respectively, thus in the same way as we've proven it for path's we construct $$H(x,t)=\{\begin{array}{cc}{H}_1(x,2t)& \text{ if }t\le \frac{1}{2}\\ H_2(x,2t-1)& \text{ if }t\ge \frac{1}{2}\end{array}$$ which remains continuous by the pasting lemma

$⟹$ Suppose that $X$ is simply connected, and suppose that ${\gamma }_a$ and ${\gamma }_b$ are paths with shared ends, both starting and ending at $x,y\in X$ respectively, thus we can conclude that ${\gamma }_a\cdot {\bar{\gamma }}_b\sim e$, but now we observe the following chain of path homotopies: $$\begin{array}{cc}{\gamma }_a& \sim {\gamma }_a\cdot ({\bar{\gamma }}_b\cdot {\gamma }_b)\\ & \sim ({\gamma }_a\cdot {\bar{\gamma }}_b)\cdot {\gamma }_b\\ & \sim {\gamma }_b\end{array}$$ thus by transitivity we know that ${\gamma }_a\sim {\gamma }_b$

$⟸$ Now suppose that $\left[\gamma \right]\in {\pi }_1(X,p)$, then $\gamma$ is closed ( a loop) , so that $\gamma \left(0\right)=\gamma \left(1\right)=p$, but also $e_p$ has this property, thus $e_p\sim \gamma$ so that $\left[\gamma \right]=\left[e_p\right]=e$ as needed.

$1\Rightarrow 2$ Suppose $f:S^1\to X$. By hypothesis, there's a homotopy $H:S^1\times I\to$ $X$ from $f$ to a constant map. That is, $H(s,0)=f\left(s\right)$ and there is a point $x\in X$ such that, for all $s\in S^1,H(s,1)=x$. Because of the latter condition, $H$ factors through the quotient $S^1\times I/S^1\times \{1\}$, which is to say there exists a function $\tilde{H}:S^1\times I\setminus S^1\times \left\{1\right\}\to X$ such that $H=\tilde{H}\circ q$ where the $q$ is the quotient map.

Note that the pair $(S^1\times I/S^1\times \{1\},S^1\times \{0\})$ is homeomorphic to $(D^2,S^1)$, the homeomorphism is given by the map $\varphi :S^1\times I\to D^2$ defined by $\varphi (a,x)=a\cdot (1-x)$, where we look at $a$ and $x$ as complex numbers. Therefore $\tilde{H}$ gives a map $D^2\to X$ such that the restriction to $S^1$ is equal to $f$

$2\Rightarrow 3$ Suppose that every map $\lambda :S^1\to X$ extends to a map ${\lambda }^{\prime }:D\to X$, now let $p\in X_0$ and suppose that $\gamma$ is a loop based at $p$, since the mapp $f\left(x\right)=e^{2\pi ix}$ is a quotient map, then we obtain some $\tilde{\gamma }:S^1\to X$ such that $\gamma =\tilde{\gamma }\circ f$, moreover $\tilde{\gamma }$ is continouous, and therefore it extends to a map ${\hat{\gamma }}^{\prime }:D\to X$.

Since $D$ is a convex set, then we may construct the straight line homotopy from the constant loop $c_p$ and the loop $f:[0,1]\to S^1\subseteq D$ where we consider $f$ as a function into $D$. Denote this homotopy as $H:I\times I\to D$ and consider the composite map ${\tilde{\gamma }}^{\prime }\circ H:I\times I\to X$ $$\begin{array}{cc}& {\tilde{\gamma }}^{\prime }\circ H(s,0)=\tilde{\gamma }\circ H(s,0)=\tilde{\gamma }\circ c_p(s)=p,\\ & {\tilde{\gamma }}^{\prime }\circ H(s,1)=\tilde{\gamma }\circ H(s,1)=\tilde{\gamma }\circ f(s)=\gamma (s),\\ & {\tilde{\gamma }}^{\prime }\circ H(0,t)=\tilde{\gamma }\circ H(0,t)=\tilde{\gamma }(1)=p,\\ & {\tilde{\gamma }}^{\prime }\circ H(1,t)=\tilde{\gamma }\circ H(1,t)=\tilde{\gamma }(1)=p,\end{array}$$ and is therefore a path homotopy from $c_p$ to $\gamma$.

$3\Rightarrow 1$ We assume that there is some $x_0\in X$ such that ${\pi }_1(X,x_0)=\left\{e\right\}$, now suppose that $\lambda :S^1\to X$ is continous, we want to show that it is a homotopic to a constant function. If we consider the continuous composition: $\lambda \circ f:[0,1]\to X$ where $f\left(x\right)=e^{2\pi ix}$ then it is a loop in $X$ and is therefore homotopic to the constant path $c$. We also know that $f\times \mathrm{id}:I\times I\to S^1\times I$ is a quotient map by the closed map theorem since it is closed surjective and continuous. Thus we obtain a map $\tilde{H}:S^1\times I\to X$ which is a homotopy between $\lambda$ and the constant map sending $S^1$ to $x_0$

To see why we note that $H(s,t)=\tilde{H}(f\left(s\right),t)$, and then it can be shown that the equation $\tilde{H}(s,t)=H(f^{-1}\left(\left\{s\right\}\right),t)$ (noting that all images and pre-images only ever return sets with one element), so we have $\tilde{H}(x,0)=H(f^{-1}\left(\left\{x\right\}\right),0)=\lambda \circ f\left(f^{-1}\left(\left\{x\right\}\right)\right)=\lambda \left(x\right)$, similarly the other equations hold true.

We first make the observation that $S^2$ can be covered by 6 open half-spheres $\{U_1,U_2,\dots ,U_6\}$ placed onto the sphere as if it was a cube with 6 sides. If we set $h\left(x\right)=e^{i2\pi x}$ then we see that $G:=\lambda \circ h:[0,1]\to S^2$ and that $\left\{G^{-1}\left(U_i\right)\right\}$ is an open cover of $[0,1]$, since it is compact, then by the lebesgue number lemma there exists some $\delta >0$ such that for each subset of $[0,1]$ having length less than $\delta$ there is an element in the open cover containing it.

So now we make an equally spaced partition of $[0,1]$ of size $n$ chosen large enough ie $\frac{1}{n}<\delta$ so that each sub-interval has size less than $\delta$, and let these intervals be denoted by $I_1,I_2,\dots ,I_n$, again by the lebesgue number lemma, for any such interval $I_j$ we know that there exists a $G^{-1}\left(U_m\right)$ (where $m\in [6]$) that contains $I_j$ completely. On the other hand this implies that $G\left(I_j\right)=\lambda \circ h\left(I_j\right)\subseteq U_m$.

Firstly note that each half sphere is homeomorphic with ${ℝ}^2$ which is simply connected. Now can think of $G{|}_{I_j}$ as being its own path from the sub-interval into $S^2$ and with the above fact, then we know this path along the surface of $S^2$ with endpoints $a,b\in U_m$, as it is simply connected, then we know that the path of the great circle connecting $a,b$ is contained within $U_m$ and that the great circle path and the sub-interval path are homeomorphic.

By repeating this process for each $I_k$ we obtain a sequence of great circle paths connecting each of the endpoints, by concatenating them we obtain a path $C[0,1]\to S^2$ that is homotopic to the path given by $G=\lambda \circ h$, but it should be clear that a finite union of great circle paths can never cover $S^2$. Since $C:[0,1]\to S^1$ and $h:[0,1]\to S^1$ is a quotient map then we obtain some function $c:S^1\to S^2$ such that $c\circ h=C$, in other words we know that $c\circ h$ and $\lambda \circ h$ are path homotopic, showing that $c$ and $\lambda$ are homotopic as functions.

But since $C\left([0,1]\right)=c\circ h\left([0,1]\right)$ which is equal to the path which is a finite number of great circle segments concatenated which we already found out does not cover $S^2$, then also since $h$ is surjective then $c\circ h(\infty 0,1)=c\left(S_1\right)$ showing that $c$ is not surjective, thus $\lambda$ is homotopic to a function $c:S^1\to S^2$ which is not surjective as needed.

We start by proving that given a continuous function ${\lambda }_0:S^1\to S^2$ that is not surjective is homotopic to a constant function, since ${\lambda }_0$ is not surjective, then there exists a point $p$ that is never reached by ${\lambda }_0$. Now we know that $S^2\setminus \left\{p\right\}$ is homeomorphic to ${ℝ}^2$ via the stereographic projection $\varphi$ and that ${ℝ}^2$ is simply connected, if we take $f\left(t\right)=e^{2piit}$ then we know that $\lambda \circ f:[0,1]\to S^2$ is a loop, and is thus homotopic to a constant path $C:[0,1]\to S^2$ given by the path homotopy $H$. Since $f\left(x\right)$ is a quotient map as noted earlier so we obtain some $c:S^1\to S^2$ such that $C=c\circ f$ so that we know the paths $\lambda \circ f$ and $c\circ f$ are homotopic, since $f$ is surjective, then we can conclude that $c$ itself is constant, and also then that $\lambda$ and $c$ are homotopic, as needed.

$⟸$ Suppose that $U\in OC$, since it is evenly covered by $p$ then there exists a collection $\{V_{\alpha }:\alpha \in I\}$ of disjoint sets whose union equals $p^{-1}\left(U\right)$, and lets construction a function $f:{⨆}_{\alpha \in I}{V}_{\alpha }\to I$ such that $f\left(p\right)=\beta$ if $p\in V_{\beta }$. Given any $\gamma \in I$ then we know that $f^{-1}\left(\gamma \right)=V_{\gamma }$ which is an open set, therefore $f$ is a continuous function.

If we construct $\varphi :{⨆}_{\alpha \in I}{V}_{\alpha }\to U\times I$ via $\varphi \left(x\right)=(p\left(x\right),f\left(x\right))$ then we know that $\varphi$ is a continuous function as the product of continuous functions, and is invertible given by ${\varphi }^{-1}$ which maps $(y,i)$ to the unique point $x\in V_{\beta }$ wherein $p\left(x\right)=y$. Moreover we know that ${\varphi }^{-1}$ is continuous because if $W$ is an open set in $U\times I$ then $\varphi \left(W\right)$ is the disjoint union of $p(W\cap V_{\alpha })\times \left\{\alpha \right\}$

$⟹$ Note that the set $p^{-1}\left(U\right)={\varphi }^{-1}(U\times I)$ which is a disjoint union of open sets $V_i={\varphi }^{-1}(U\times \left\{\alpha \right\})$, each of which is mapped homoeomorphically onto $U$.

$⟹$ Suppose that there is a lift $\tilde{f}:X\to E$, that is: $f=p\circ \tilde{f}$, we'll now prove the inclusion, to do so we consider an element $f_{\ast }\left[\gamma \right]$, where $\gamma$ is a loop in $X$ based at $x_0$, then note: $$f_{\ast }\left[\gamma \right]={(p\circ \tilde{f})}_{\ast }\left[\gamma \right]=p_{\ast }\left(f_{\ast }\left[\gamma \right]\right)$$ but now we note that $f_{\ast }\left[\gamma \right]$ is a loop in $E$ based at $e_0$, therefore $p_{\ast }\left(f_{\ast }\left[\gamma \right]\right)\in p_{\ast }\left({\pi }_1(E,e_0)\right)$ as needed.

$⟸$ Now we assume that the inclusion holds true and now we must construct a lift, $\tilde{f}:(X,x_0)\to (B,b_0)$, so we know that $\tilde{f}\left(x_0\right)=b_0$ must be the case, but we still must define this function for all other different points in $X$, so let $x\in X$ be such a point, then because $X$ was assumed to be path connected, then we obtain a path $\alpha [0,1]\to X$ from $x_0$ to $x$, now if we consider $f\circ \alpha :I\to B$ which is a path in $B$ starting at $b_0=f\left(x_0\right)$ and ending at $f\left(x\right)$ [1: this fact will become important later on] then by the path lifting lemma we obtain a path ${\widetilde{f\circ \alpha }}_x:I\to E$ such that ${\widetilde{f\circ \alpha }}_x\left(0\right)=e_0$, for reasons that will come to be clear we now define $\tilde{f}\left(x\right)={\widetilde{f\circ \alpha }}_x\left(1\right)$

We must now verify that the equation we just wrote above defining $\tilde{f}$ must lead to an actual function, that is continuous, that it is a lift of $f$ and moreover that it is the unique one satisfying $\tilde{f}\left(x_0\right)=e_0$.

To verify that $\tilde{f}$ is a lift, we must check that $p\circ \tilde{f}=f$, but for any $x\in X$ we have that $$p\circ \tilde{f}\left(x\right)=p\left(\tilde{f}\left(x\right)\right)=p\left({\widetilde{f\circ \alpha }}_x\left(1\right)\right)=f\left(\alpha \left(1\right)\right)=f\left(x\right)$$ ie, $p\circ \tilde{f}=f$, as needed.

To show the uniqueness of $\tilde{f}$ we don't have to do much, because as we set it equal to the lift $\widetilde{f\circ \alpha }$ and in the path lifting lemma we showed that it was a unique lift, then it implies that our constructed $\tilde{f}$ is also unique.

We must also show that $\tilde{f}$ is well defined because it was defined in terms of the first path we got from $x_0$ to $x$ since $X$ was path connected, but since there could be many such resulting paths from $x$ to $x_0$ we must verify that on these paths $\tilde{f}$ outputs the same value. So let $x\in X$ and suppose that $\alpha ,\beta$ are two paths in $X$ from $x_0$ to $x$, we must verify that $$\widetilde{f\circ \alpha }\left(1\right)=\tilde{f}\left(x\right)=\widetilde{f\circ \beta }\left(1\right)$$ Since $\alpha ,\beta$ have shared ends and both end at $x$ then we can conclude that $f\left(\alpha \left(1\right)\right)=f\left(x\right)=f\left(\beta \left(1\right)\right)$, then the loop $\gamma :={\beta }^{-1}\cdot \alpha$ is a loop in $X$ based at $x_0$, and therefore $f\circ \gamma$ is a loop in $E$ based at $e_0$, since we assumed that $f_{\ast }\left({\pi }_1(Y,y_0)\right)\subset p_{\ast }\left({\pi }_1(\tilde{X},x_0)\right)$, which implies that that there is a loop $\delta$ in $B$ based at $b_0$ such that $p\circ \delta =f\circ \gamma$.

We'll prove that $\widetilde{f\circ \alpha }\left(1\right)=\delta \left(\frac{1}{2}\right)=\widetilde{f\circ \beta }\left(1\right)$, to see why we see that $\delta {\upharpoonright }_{[0,\frac{1}{2}]}$ is a lift of $f\circ \alpha$ starting at $b_0$, since by the path lifting lemma since it deduces that the lift is unique then we must have that $\widetilde{f\circ \alpha }=\gamma {\upharpoonright }_{[0,\frac{1}{2}]}$ so we must have that $\delta \left(\frac{1}{2}\right)=\widetilde{f\circ \gamma }\left(1\right)=f\left(x\right)$. In a symmetrical way we can see that ${\delta }^{-1}{\upharpoonright }_{[0,\frac{1}{2}]}$ is a lift of $\widetilde{f\circ \beta }$ starting at $b_0$ so similarly we conclude that ${\delta }^{-1}\left(\frac{1}{2}\right)$, but since $\delta \left(\frac{1}{2}\right)={\delta }^{-1}\left(\frac{1}{2}\right)$ then that shows that $\widetilde{f\circ \alpha }=\widetilde{f\circ \beta }$, ie that $\tilde{f}$ is well defined.

Finally we will prove that $\tilde{f}$ is continuous, so let $U\subseteq E$ be an open set and we need to prove that ${\tilde{f}}^{-1}\left(V\right)$ is open. To do this we will show that for any $x\in {\tilde{f}}^{-1}\left(U\right)$ there exists an open set $V$ such that $x\in V\subseteq f^{-1}\left(U\right)$ showing that $f^{-1}\left(U\right)$ is open

Since $p$ is a covering, then recall that this means that we have an open cover $OC$ of $B$ with various properties, in our case if we consider $f\left(x\right)$ then we obtain an open set $W$ that is evenly covered by $p$ and consider the set $f^{-1}\left(W\right)\subseteq X$ which is open as $f$ is continuous and since $X$ is locally path connected then there exists an open set $V$ such that $x\in V\subseteq f^{-1}\left(W\right)$ and $V$ is path connected.

Now fix a path $\gamma$ from $x_0$ to $x$ and for any $v\in V$ let $\eta$ be a path from $x$ to $v$ both of which can be done as $X$ was path connected, then $\gamma \cdot \eta$ is a path from $x_0$ to $v$, and $f\circ (\gamma \cdot \eta )$ is a path in $B$ starting at $f\left(x\right)$ and ending at $f\left(v\right)$, since $p$ was a homoemorphism then we also have $p^{-1}$ so that $p^{-1}\circ f\circ (\gamma \cdot \eta )$ is a path in $E$ that starts at $\tilde{f}\left(x\right)$ and ends at $\tilde{f}\left(v\right)$ which is contained in $U$ , in other words ${\tilde{f}}^{-1}\left(U\right)\supseteq W$ as needed.

Recall that ${\pi }_X:X\times Y\to X$, ${\pi }_Y:X\times Y\to Y$ are the projections onto $X$ and $Y$ respectively. Now to show that these two groups are isomorphic we construction a function $\varphi :{\pi }_1(X\times Y,(x_0,y_0))\to {\pi }_1(X,x_0)\times {\pi }_1(Y,y_0)$, we do it by taking $\left[\gamma \right]\mapsto ([{\pi }_X\circ \gamma ],[{\pi }_Y\circ \gamma ])$, also do not forget that $\gamma$ is a loop based at $(x_0,y_0)$.

First we have to prove that $\varphi$ is well-defined, that is if $\left[\alpha \right]=\left[\gamma \right]$ where $\alpha$ is another loop in $X\times Y$ based at $(x_0,y_0)$ then we need to show that $([{\pi }_X\circ \alpha ],[{\pi }_Y\circ \alpha ])=([{\pi }_X\circ \gamma ],[{\pi }_Y\circ \gamma ])$.

By assuming that $\left[\alpha \right]=\left[\gamma \right]$ we know that $\alpha \sim \gamma$ meaning that they are path homotopic, moreover since ${\pi }_X,{\pi }_Y$ are continuous, then we know that ${\pi }_X\circ \gamma \sim {\pi }_X\circ \alpha$ and that ${\pi }_Y\circ \gamma \sim {\pi }_Y\circ \alpha$ are also path homotopic, which shows that $([{\pi }_X\circ \alpha ],[{\pi }_Y\circ \alpha ])=([{\pi }_X\circ \gamma ],[{\pi }_Y\circ \gamma ])$ as needed.

We construct ${\varphi }^{-1}$ (which will be the inverse) is defined by ${\varphi }^{-1}(\left[\gamma \right],\left[\alpha \right])=\left[(\gamma ,\alpha )\right]$ where $(\gamma ,\alpha )\left(t\right)=(\gamma \left(t\right),\alpha \left(t\right))$ is a loop in $X\times Y$. So we also have to verify that this is well defined, so suppose that $\left[\tilde{\alpha }\right]=\left[\alpha \right]$ and that $\left[\tilde{\gamma }\right]=\left[\gamma \right]$ then we have to prove that $\left[(\gamma ,\alpha )\right]=\left[(\tilde{\gamma },\tilde{\alpha })\right]$. Since we assumed that $\gamma \sim \tilde{\gamma }$ and that $\alpha \sim \tilde{\alpha }$ have have path homotopies $H_1,H_2$ for each pair respectively, then if we consider $(H_1(s,t),H_2(s,t))(s,t)$ it is a path homotopy from $(\gamma ,\alpha )$ to $(\tilde{\gamma },\tilde{\alpha })$ as needed.

Now $\varphi$ is a homomorphism, because we have $$\begin{array}{cc}\varphi (\left[\gamma \right]\cdot \left[\alpha \right])& =\varphi \left([\gamma \cdot \alpha ]\right)\\ & =([{\pi }_X\circ (\gamma \cdot \alpha )],[{\pi }_Y\circ (\gamma \cdot \alpha )])\\ & =([{\pi }_X\circ \gamma ]\cdot [{\pi }_X\circ \alpha ],[{\pi }_Y\circ \gamma ]\cdot [{\pi }_Y\circ \alpha ])\\ & =([{\pi }_X\circ \gamma ],[{\pi }_Y\circ \gamma ])\cdot ([{\pi }_X\circ \alpha ],[{\pi }_Y\circ \alpha ])\\ & =\varphi \left(\left[\gamma \right]\right)\cdot \varphi \left(\left[\alpha \right]\right)\end{array}$$