So in this proof, that the fundamental group $\pi _1 (X, x_0) $ is a group, it involves constructing a homotopy from $f*e$ to $f$ and $f*e$ to $e*f$ and they say to draw these diagrams to get an intuition of how to construct the homotopies. I just can’t figure out what’s going on the the diagrams and how I use them to describe the homotopy. Can anyone give me an idea of what’s going on here?
This is from an old Topology Explained note that has sadly disappeared from the internet, so I repost it here, thx to Elliott Pearl who TeXed my original plain text answer on "ask a topologist"".
Intro
We have a space $X$ and point $p$ in it. Consider the set $Y$ of closed paths beginning and starting at $p$: these are all continuous maps $f\colon[0,1] \to X$ such that $f(0)=f(1)=p$. On this set we have the equivalence relation of being homotopic: $f$ and $g$ in $Y$ are homotopic when there is a continuous map $H\colon [0,1]\times [0,1] \to X$ such that
Such a map $H$ is called a homotopy between $f$ and $g$, and we will write $f \simeq g$ when such an $H$ exists. $H(s,t)$, as $t$ varies, is a deformation (a path) from $f(s)$ to $g(s)$, and this for all $s$ simultaneously and continuously. Condition (3) is to make sure that all intermediate maps $H(\cdot,t)$ are all closed paths from $p$ (these are also called loops centered at $p$). This is an equivalence relation on $Y\colon f \simeq f$ by the homotopy $H(s,t) = f(s)$. If $f \simeq g$ by $H$, then $g \simeq f$ by the homotopy $H'$ that sends $(s,t)$ to $H(s,1-t)$. We reverse time, in a way \dots And to check transitivity, if $f \simeq g$ by $G$ and $g \simeq h$ by $H$ then we get $f \simeq h$ by the homotopy $$ F(s,t) = \begin{cases} G(s,2t)&\text{if $0 \le t \le \frac12$, and}\\ H(s,2t-1)&\text{if $\frac12 \le t \le 1$.} \end{cases} $$ Then $F$ is well-defined because $F(s,\frac12) = G(s,1) = g(s)$ by the upper definition while the other definition for $t=\frac12$ gives us $H(s,0) = g(s)$ as well. On $[0,1] \times [0,\frac12]$ the map $F$ is continuous (as $G$ is) and on $[0,1] \times [\frac12,1]$ the map $F$ is continuous as well (as $H$ is), so standard facts about continuous maps defined on finitely many closed sets gives us that $F$ is also continuous. Moreover $F(s,0) = G(s,0) = f(s)$; $F(s,1) = H(s,1) = h(s)$; $F(0,t)$ is either $G(0,2t)$ or $H(0,2t-1)$ which is always $p$. So $F$ is indeed the required homotopy. This shows that being homotopic is indeed an equivalence relation on $Y$, the set of loops centered at $p$. The class of $f$ will be denoted $[f]$, and the set of all equivalence classes is denoted $\pi_1(X,p)$. We will show next that this set can be made into a group, and it is called the first homotopy group of $X$ at $p$.
What will the operation $\ast$ on $\pi_1(X,p)$ be? It will be composition of paths: if $f$ is a loop around $p$, and $g$ is another one then we can define a loop $f \ast g$ as follows: follow $f$ at double the speed (so for $t=0$ to $\frac12$) and then g at double the speed, from $\frac12$ to $1$. In a formula: $f \ast g\colon [0,1] \to X$, where $$ (f \ast g)(t) = \begin{cases} f(2t)&\text{for $t$ in $[0,\frac12]$,}\\ g(2t-1)&\text{for $t$ in $[\frac12,1]$.} \end{cases} $$ Then $f \ast g$ is defined on two closed sets, and their definition concides for the intersection $t=\frac12$: $f(2 \cdot \frac12) = f(1) = p$ while $g(2 \cdot \frac12 -1) = g(0) = p$ as well. And $f$ and $g$ are continuous, so $t \mapsto f(2t)$ and $t \mapsto g(2t-1)$ are as well (compositions of the maps $t \mapsto 2t$ and $f$, and $t \mapsto 2t-1$ and $g$ respectively). Moreover, $(f \ast g)(0) = f(0) = p$ and also $(f \ast g)(1) = g(2 \cdot 1 - 1) = g(1) = p$, so that $f \ast g$ is indeed a loop.
But all this is true for loops. We have to define all this on $\pi_1(X,p)$, which is a set of equivalence classes, under homotopy. There is only one way to go: let $[f]$ and $[g]$ be in $\pi_1(X,p)$, the classes of $f$ and $g$ respectively. Define $[f] \ast [g]$ to be $[f \ast g]$, the class of the composition loop of $f$ and $g$. This seems to depend on the representatives, so we have to show it is well-defined: if $f'$ is in $[f]$ and $g'$ is in $[g]$ then $f' \ast g'$ is in $[f \ast g]$. Or put slightly differently: if $f \simeq f'$ (say by $H$) and $g \simeq g'$ (say, by $G$), then we have to show that $f \ast g \simeq f' \ast g'$, by some (yet to be constructed) homotopy $F$. The idea is quite easy: $F$ has to follow $H$ for the first half of the domain, and $G$ for the second, mimicking the definition of composition of loops: $$ F(s,t) = \begin{cases} H(2s,t)&\text{if $(s,t)$ is in $[0,\frac12] \times [0,1]$,}\\ G(2s-1,t)&\text{if $(s,t)$ is in $[\frac12,1] \times [0,1]$.} \end{cases} $$ $F$ is continuous on both closed parts (quite obvious) and the overlap is for $s=\frac12$, $t$ arbitary. In that case $F(\frac12,t) = H(1,t)$ on the one hand, and $F(\frac12,t) = G(0,t)$ on the other. But both $H$ and $G$ are homotopies of loops, so they keep $p$ fixed at all times, so both sides are always equal to $p$. So $F$ is well-defined and continuous. \begin{align*} F(s,0)& = H(2s,0)&&\text{(for $\textstyle s \le \frac12$)}\\ & = f(2s,0)&&\text{(as $H$ is an homotopy between $f$ and $f'$)}\\ & = (f \ast f')(s)&&\text{for such $s$.} \end{align*} For $s \le \frac12$ we also have $F(s,1) = H(2s,1) = f'(2s) = (f' \ast g')(s)$. For $\frac12 \le s \le 1$ we have $F(s,0) = G(2s-1,0) = g(2s-1) = (f \ast g)(s)$ and $F(s,1) = G(2s-1,1) = g'(2s-1) = (f' \ast g')(s)$ (for such $s$). So for all $s$: $F(s,0) = (f \ast g)(s)$ and $F(s,1) = (f' \ast g')(s)$. Moreover $F(0,t) = H(0,t) = p$ for all $t$ and also $F(1,t) = G(1,t) = p$ for all $t$, so $F$ keeps the endpoints at $p$ for all $t$, as required by part (3) of the definition of homotopy of loops. As you see, it's quite a bit of writing out if we want all the details...
Group axioms
Having established that $\pi_1(X,p)$ has an operation $\ast$ on it, we would like it to be a group. The first thing to show is associativity, so for all $[f],[g],[h]$ in $\pi_1(X,p)$ we want to have that $[f] \ast ([g] \ast [h]) = ([f] \ast [g]) \ast [h]$. Or, working on the representatives $f$,$g$ and $h$: $f \ast (g \ast h) \simeq (f \ast g) \ast h$, for loops $f,g,h \colon [0,1] \to X$ at $p$. So what do these maps $f \ast (g \ast h)$ and $(f \ast g) \ast h$ look like? The first is the result of first composing $g$ and $h$ and then composing with $f$ in front of it. So on $0 \le s \le \frac12$ this map is equal to $f(2s)$. On $0 \le s \le \frac12$ it is equal to $(g \ast h)(2s-1)$, which has to be split again: for $0 \le 2s-1 \le \frac12$, or $\frac12 \le s \le \frac34$, we have $(g \ast h)(2s-1) = g(2 \cdot (2s-1)) = g(4s-2)$, while for $\frac34 \le s \le 1$ we have $(g \ast h)(2s-1) = h(2 \cdot (2s-1) - 1)) = h(4s-3)$. So in short, $$ f \ast (g \ast h)(s) = \begin{cases} f(2s)&\text{for $s$ in $[0,\frac12]$,}\\ g(4s-2)&\text{for $s$ in $[\frac12,\frac34]$,}\\ h(4s-3)& \text{for $s$ in $[\frac34,1]$.} \end{cases} $$
What about $(f \ast g) \ast h$? Here we first composed $f \ast g$ in the usual way, and then composed the result with $h$. This means that for $s \ge \frac12$ we have that this map equals $h(2s-1)$.
In the case $s \le \frac12$ we have to split again: at first we see that $(f \ast g) \ast h$ equals $(f \ast g)(2s)$ here, so when $2s \le \frac12$ we have that this equals $f(2 \cdot 2s) = f(4s)$, while $\frac12 \le 2s \le 1$, or $\frac14 \le s \le \frac12$, gives the result $g(2 \cdot 2s - 1) = g(4s-1)$. In short again, $$ (f \ast g) \ast h (s) = \begin{cases} f(4s)&\text{for $s$ in $[0,\frac14]$,}\\ g(4s-1)&\text{for $s$ in $[\frac14,\frac12]$,}\\ h(2s-1)&\text{for $s$ in $[\frac12,1]$.} \end{cases} $$
So in fact both maps trace out $f$, then $g$, and finally $h$ but in different speeds during the interval $[0,1]$. How to find a homotopy? One way to look at it, is to take the square $[0,1] \times [0,1]$ (coordinates $(s,t)$) with two line segments in it: $l_1$ from $(\frac14,0)$ to $(\frac12,1)$, $l_2$ from $(\frac12,0)$ to $(\frac34,1)$. At the top of the square $l_1$ and $l_2$ divide up the domain $[0,1] \times \{1\}$ like $f \ast (g \ast h)$ does $(0\text{--}\frac12\text{--}\frac34\text{--}1)$, while at the bottom they divide up the domain $[0,1] \times \{0\}$ as $(f \ast g) \ast h$ does. Now we contract the square down to the bottom line $[0,1] \times \{0\}$ continuously, and in the mean time we make sure that on each intermediate stage $[0,1] \times \{t\}$ we let the homotopy trace $f$ on the leftmost part: from the point $(0,t)$ to the intersection point of $l_1$ with $[0,1] \times \{t\}$; it traces $g$ on the middle part (between $l_1$ and $l_2$), and $h$ on the rightmost part, up to $(1,t)$ . The equation of $l_1$ is $t = 4 \cdot s - 1$, as can easily be checked. The equation of $l_2$ is $t = 4 \cdot s - 2$. So the leftmost part on level t is given by an interval that starts with $s=0$ and goes on to $s = \frac{t+1}{4}$ (solve $s$ from the equation for $l_1$). So this means that for $s$ in $[0,4t+1]$ we have to define $H(s,t)$ to be $f(\frac{4s}{t+1})$: at $s=0$ we have $f(0)$, and it increases in $s$ up to $f(\frac{4 \cdot \frac{t+1}{4}}{{t+1}}) = f(1)$. The middle part is delimited by $s = \frac{t+1}{4}$ up to $\frac{t+2}{4}$ (also solve $s$ from the equation for $l_2$).
So we have to find a linear (for ease) formula $\phi$ in $s$ and $t$ such that $\phi(\frac{t+1}{4}, t) = 0$ and $\phi(\frac{t+2}{4}, t) = 1$. Trying out $\phi(s,t) = a s + b t + c$, for some $a$,$b$,$c$ and substituting, we find that $a=4$, $b=-1$ and $c=-1$ works. So we define $H(s,t)$ for this combination of $s$ and $t$ to be equal to $g(4s-t-1)$. For fixed $t$, this increases nicely from $0$ to $1$ when $s$ goes from $\frac{t+1}{4}$ to $\frac{t+2}{4}$, so that $g$ traces its normal trajectory exactly in this interval. Finally, on the rightmost part, so for fixed $t$, $s$ varying from $s = \frac{t+2}{4}$ to $1$, we want another $\psi(s,t)$ such that $\psi(\frac{t+2}{4}, t) = 0$ and $\psi(1,t) = 1$, as above. A linear map in $s$ (fixed $t$) that sends $\frac{t+2}{4}$ to $0$ and $1$ to $1$ is $\frac{4}{2-t} \cdot s + \frac{-2-t}{2-t}$ (check this!). So (simplifying this by putting $\frac{1}{2-t}$ outside brackets) we get that for the relevant $s$ and $t$ we must put $H(s,t)$ as $h(\frac{4s-t-2}{2-t})$. So summarising: $$ H(s,t) = \begin{cases} f(\frac{4s}{t+1})& \text{for $s$ in $[0, \frac{t+1}{4}]$,}\\ g(4s -t -1)& \text{for $s$ in $[\frac{t+1}{4}, \frac{t+2}{4}]$,}\\ h(\frac{4s-t-2}{2-t})& \text{for $s$ in $[\frac{t+2}{4}, 1]$.} \end{cases} $$
We have thus defined $H$ for all of $[0,1] \times [0,1]$. As to the boundaries between the areas (delimited by the lines $l_1$ and $l_2$) we see, e.g., that for $s = \frac{t+1}{4}$ the first part of the definition gives $f(1)$, and the second $g(0)$ and both are equal to $p$. Similarly for the boundary $s = \frac{t+2}{4}$, where we get $g(1) = h(0) = p$. So $F$ is well-defined (no conflicts on boundaries) and continuous as it is on each part. For $t=0$ we get $f(\frac{4s}{1}) = f(4s)$ on $[0,\frac14]$, $g(4s-2)$ on $[\frac14,\frac12]$ (again fill in $t=0$) and $h(\frac{4s-2}{2-0}) = h(2s-1)$ for $s = [\frac12,1]$, hence we get exactly $((f \ast g) \ast h)(s)$ as $H(s,0)$. Taking $s=1$ we get $f(\frac{4s}{1+1}) = f(2s)$ on $[0,\frac12]$, $g(4s - 1 - 1) = g(4s-2)$ for $s$ in $[\frac12,\frac34]$, and $h(\frac{4s-3}{2-1}) = h(4s-3)$ for $s$ in $[\frac34,1]$. So this gives exactly $(f \ast (g \ast h))(s)$. Finally, for all $t$: $H(0,t) = f(0) = p$ (via case 1) and $H(1,t) = h(\frac{4-t-2}{2-t}) = h(\frac{2-t}{(2-t}) = h(1) = p$ (via case 3). Conclusion: $H$ is indeed a homotopy between $(f \ast g) \ast h$ and $f \ast (g \ast h)$, as required. This gets rid of the trickiest part, the associativity.
There is another way to prove that functions are homotopic: via the notion of reparametrisation. If $f$ is a loop at $p$, then a reparametrisation of $f$ is a map $g$ of the form $g = f \circ \phi$, where $\phi\colon [0,1] \to [0,1]$ is a continuous map such that $\phi(0) = 0$ and $\phi(1) = 1$. This is again a loop at $p$: continuity follows from that of $f$ and $\phi$, while $g(0) = f(\phi(0)) = f(0) = p$ and $g(1) = f(\phi(1)) = f(1)= p$ as well. Moreover $f$ and $g = f \circ \phi$ are homotopic: define $H_{\phi}\colon [0,1] \times [0,1] \to X$ by $H_{\phi}(s,t) = f((1-t) \cdot \phi(s) + t \cdot s)$. This is well-defined, as $(1-t) \cdot \phi(s) + t \cdot s$ lies between $\phi(s)$ and $s$ so that it is a point in $[0,1]$, and $f$ is defined on it. Also $H_{\phi}(s,0) = f(\phi(s)) = g(s)$, and $H_{\phi}(s,1) = f(s)$. Moreover: $$H_{\phi}(0,t) = f((1-t) \cdot \phi(0) + t \cdot 0) = f(0) = p,$$ using $\phi(0)=0$. And $$H_{\phi}(1,t) = f((1-t) \cdot \phi(1) + t \cdot 1) = f(1) = p$$ using $\phi(1) =1$. So $H_{\phi}$ is a homotopy between $f$ and $g$. Now, given $f$,$g$ and $h$ (all loops at $p$), we consider the map $\phi \colon [0,1] \to [0,1]$ given by: $$\phi(s) = \begin{cases} \frac12 \cdot s&\text{for $s$ in $[0,\frac12]$,}\\ s - \frac14&\text{for $s$ in $[\frac12,\frac34]$,}\\ 2 \cdot s - 1&\text{for $s$ in $[\frac34,1]$.} \end{cases} $$
The graph is a broken line from $(0,0)$ via $(\frac12,\frac14)$ and $(\frac34,\frac12)$ to $(1,1)$. (Draw this!) Look at $((f \ast g) \ast h)(\phi(s))$: for $s$ in $[0,\frac12]$ this equals $(f \ast g) \ast h (\frac12 \cdot s) = f(4 \cdot \frac12 \cdot s) = f(2s)$ (as $\frac12 \cdot s$ in is $[0,\frac14]$, see the formula for $(f \ast g) \ast h$ above). For $s \in [\frac12,\frac34]$ we have that $\phi(s) = s - \frac14$ in $[\frac14,\frac12]$, so $((f \ast g) \ast h)(\phi(s)) = g(4(s - \frac14) - 1) = g(4s-2)$, and for $s$ in $[\frac34,1]$ we see that $\phi(s) = 2s -1$ in $[\frac12,1]$, so $((f \ast g) \ast h)(\phi(s)) = h(2(2s-1) - 1) = h(4s-3)$. So $(f \ast g) \ast h \circ \phi = f \ast (g \ast h)$ (check the formula for the latter function above!) so that $f \ast (g \ast h)$ is a reparametrisation of $(f \ast g) \ast h$, and as such they are homotopic, as we saw.
What about the identity? Define $e_p$ to be the constant loop: $e_p(s) = p$ for all $s$ in $[0,1]$. We claim that $[e_p]$ is the identity element of $\pi_1(X,p)$. Let's prove it is a left identity element, so that for all $f$: $[e_p] \ast [f] = [f]$, or $e_p \ast f \simeq f$. What is the left hand side exactly? $e_p \ast f (s) = e_p(2s) = p$ for $s$ in $[0,\frac12]$ and $e_p \ast f (s) = f(2s-1)$ for $s$ in $[\frac12,1]$.
So if we define $\phi(s) = 0$ for $s$ in $[0,\frac12]$ and
$\phi(s) = 2s-1$ for $s$ in $[\frac12,1]$, we see that $\phi(0) = 0$ and $\phi(1)=1$ (and $\phi$ is well-defined and continuous!) while also $f(\phi(s)) = f(0) = p$ for $s$ in $[0,\frac12]$ and $f(\phi(s)) = f(2s -1)$ for $s$ in $[\frac12,1]$. So $e_p \ast f$ is a reparametrisation of f and hence these maps are homotopic. An explicit formula can also be given for this homotopy:
$$ \begin{cases} H(s,t) = f((1-t) \cdot s) & s \in [0,\frac12]\\ H(s,t) = f(st-t+s) & s \in [\frac12,1] \end{cases} $$ (obtained from the general homotopy between a map and its reparametrisation after substituting the concrete $\phi$). One can easily check now that this works. We'll leave the question of right identity for later (or the reader can find an argument why $[e_p]$ must also be a right identity...)
We also need inverses, of course. Again, we'll only prove left inverses. For a loop $f\colon [0,1] \to X$, we define the reverse loop $\tilde{f}:[0,1] \to X$ to be the map $\tilde{f}(s) = f(1-s)$, for all $s$ in $[0,1]$. This traverses the trajectory of $f$ in opposite direction, so intuitively, $\tilde{f} \ast f$ should be homotopic to $e_p$.
How to see that? At level $t$ (so fix $t$ for a while) the homotopy should be a loop starting at $p$ (of course) and going to $f(t)$ backwards (so $f$ in reverse direction) in the first half of the interval, while going to $p$ (following $f$ forwards again) in the second half of the interval. If we look at the square $[0,1]^2$ (coordinates $(s,t)$) with the lines $t = 1 - 2s$ and $t = 2s -1$, we see a V-shaped section. In this section (at each level $[0,1] \times \{t\}$) we let the homotopy stand still at $f(t)$, while to the left it follows $\tilde{f}$ and on the right of it, $f$. So $$ H(s,t) = \begin{cases} f(1-2s) (=\tilde{f}(2s))& \text{for $s$ in $[0,\frac{1-t}{2}]$,}\\ f(t)& \text{for $s$ in $[\frac{1-t}{2}, \frac{1+t}{2}]$,}\\ f(2s-1)& \text{for $s$ in $[\frac{1+t}{2}, 1]$.} \end{cases} $$ Then for $s = \frac{1-t}{2}$ we get $f(t)$ in both cases and likewise for $s = \frac{1+t}{2}$. At the bottom, so for $t=0$, we see that the middle interval reduces to $[\frac12,\frac12]$, so we can discard it, and we are left with $H(s,0) = \tilde{f}(2s)$ (for $s \le \frac12$) and $f(2s-1)$ for larger $s$, which is just $\tilde{f} \ast f$. For $t=1$ we get that the first and last interval shrivel down to
$[0,0]$ and $[1,1]$ respectively, so that for all $s$ we have $H(s,1) = f(1) = p = e_p(s)$. And for $s=0$ we always have $H(0,t) = f(1) = p$ and $H(1,t) = f(1) = p$ as well. So $H$ is a valid (continuous of course) homotopy between $\tilde{f} \ast f$ and $e_p$, so $[\tilde{f}] \ast [f] = [e_p]$, showing that $[\tilde{f}]$ is a left inverse of $[f]$.
Now we are done by simple group theory: a set $G$ with associative operation $\ast$ with left identity $e$ and left inverses $\tilde{g}$ for each $g$, is a group: First note that the law $g \ast g = g \Longrightarrow g = e$ holds : multiply both sides of $g \ast g = g$ with $\tilde{g}$: $\tilde{g} \ast (g \ast g) = (\tilde{g} \ast g) \ast g = e \ast g = g$ on the one hand, while $\tilde{g} \ast g = e$, so $g=e$. Then note that for each $h$: $$\begin{align} (h \ast \tilde{h}) \ast (h \ast \tilde{h}) &= h \ast (\tilde{h} \ast h) \ast \tilde{h}\quad\text{(associativity)}\\ &= h \ast e \ast \tilde{h}\\ &= h \ast \tilde{h}, \end{align}$$ so we apply the law, with $g = h \ast \tilde{h}$ to see that $h \ast \tilde{h} = e$ for all $h$. Moreover: $$\begin{align} h \ast e& = h \ast (\tilde{h} \ast h)\\ &= (h \ast \tilde{h}) \ast h\\ &= e \ast h\quad\text{(by what we just proved)}\\ &= h \end{align}$$ so $e$ is a right identity and $\tilde{h}$ is a right inverse as well.
This concludes the proof that $\pi_1(X,p)$ is a group, in all its gory details. The group theory bit is just an elegant way for me to avoid writing out two more homotopies, of course...