Algebra - 5

0 - Lebesgue Numbers

Recall that a continuous map $p\colon X \to Y$ is called a covering map if $p$ is surjective and for each $y \in Y$ there exists an open neighborhood $U_y$ that is evenly covered by $p$.

Also recall that an open $U \subset Y$ is evenly covered by $p$ if $p^{-1}(U) = \bigsqcup_{a \in A} V_a$ where each $V_a$ is open in $X$ and each $p|_{V_a}\colon V_a \to U$ is a homeomorphism.

We feel as though the evenly covered neighborhood(s) of a point are “small” in some sense. As an exaggeration to the opposite, if $Y$ itself is evenly covered, then $p$ is a homeomorphism.

But this notion of small can only be with respect to some metric space, and a map from that metric space to $Y$, which of course has to be continuous.

We may first look at compact metric spaces. Let $M$ be a compact metric space, $\mathscr{U}$ an open cover of $M$.

The rough intuition is as follows.

Each point is in some open set of the cover. Take a ball around each point, with radius small enough to still be wholly contained inside that open set (always possible, since open sets are exactly unions of such balls).

Now we have a ball around each point of $M$, each one wholly contained inside some member of the original cover.

Halve each ball’s radius, keeping the same centre, so each half-radius ball sits wholly inside its own full-radius parent ball.

The half-radius balls still cover $M$ (each point sits at the centre of its own), so take a finite subcover of these. Each ball in this finite subcover has a centre whose full-radius (unhalved) ball is still wholly contained inside some member of the original cover.

Now take the minimum of all these (finitely many) half-radii, call it $\epsilon^*/2$. For any set of diameter less than $\epsilon^*/2$, pick a point $p$ in it, it lies inside one of the finitely many half-radius balls; every other point of the set is then within $\epsilon^*/2$ of that $p$ too, so within $\epsilon^*/2 + \epsilon^*/2 = \epsilon^*$ of the centre at most, well within the full-radius parent ball, which sits inside a member of the original cover.

So any set of small enough diameter is contained in some member of the original cover.

The Lebesgue Number Theorem

Theorem. Let $M$ be a compact metric space, and $\mathscr{U}$ be an open cover of $M$, there exists $\epsilon >0$ such that whenever $A \subset M$ and $\operatorname{diam}(A) < \epsilon$, then $A \subset U$ for some $U \in \mathscr{U}$. This $\epsilon$ of course depends on the cover itself.

Proof. for each $m \in M$ take two balls (hehe) $B(m,r_m/2) \subset B(m,r_m) \subset U_m$. Now take a finite subcover $K$ of the half radius balls, then for each $1 \leq i \leq |K|$ we have $B(m_i,r_{m_i}/2) \subset B(m_i, r_{m_i}) \subset U_{m_i}$.

Now, let $e = \min_{1 \leq i \leq |K|}(r_{m_i}/2)$, this obviously exists since $K$ is a finite set.

let $A \subset M$ for which $\operatorname{diam}(A) < e$, choose any point $a \in A$. since $K$ covers $M$, we have some $m$ for which $a \in B(m,r_m/2)$. for any point $a' \in A$, $d(m,a') \leq d(m,a) + \operatorname{diam}(A) < r_m/2 + e$ but $e \leq r_m/2$, thus $d(m,a') < r_m$, therefore $A \subset B(m,r_m) \subset U_m$.

Lebesgue Numbers for Covering Maps

Now consider the setup from above, $p\colon X \to Y$ a covering map between topological spaces, and $\alpha\colon M \to Y$ a continuous map to $Y$ from a compact metric space $M$.

Theorem. There exists an $e > 0$, such that any cover of $M$ with open $e$-balls has the property that the image of a ball under $\alpha$ is contained inside an evenly covered set in $Y$.

Proof. for each $y \in Y$ construct an open cover $P$ by choosing an evenly covered neighborhood of $y$, $U_y$. the pre-images of these form an open cover $V = \{\alpha^{-1}(U_y) \mid U_y \in P\}$ of $M$, since $M$ is compact, let $e + \delta$ be the Lebesgue number for this cover. then we know that for any ball $B(m,e) \subset \alpha^{-1}(U_y)$ thus $\alpha(B(m,e)) \subset U_y$.

we shall call this $e$ a $(p,\alpha)$ Lebesgue number.

That is to say, the thus obtained Lebesgue number quantifies the smallness of these evenly covered sets, relative to that metric space and that continuous map.

1 - Compact Metric Lifting

Small Covers of Compact Metric Spaces

We will first quickly show that a compact metric space is bounded. take some $x \in M$ and construct open balls of increasing radius $B(x,r), B(x,2r),\ldots$ of course since the distance between any two points is finite, each point is contained inside some ball, now take the finite subcover of this cover, and notice that every point is contained inside the largest ball.

Next we shall show that for any $e > 0$ a compact metric space can be covered with finitely many sets of diameter less than $e$.

consider any cover of this space with balls of radius $e/4$. take its finite subcover, the diameter of these balls is less than $e/2$, replace each ball with some $e-\delta$ diameter subset that contains the ball.

The Compact-Metric Lifting Theorem

We shall use this to solve the compact-metric lifting problem.

let $M$ be a compact metric space, $p\colon X \to Y$ a covering map, $\alpha\colon M \to Y$ a continuous map that needs to be lifted.

Theorem. The solution exists. Moreover, if for some $m_0 \in M$, we want $\tilde{\alpha}(m_0) = x_0 \in p^{-1}(\alpha(m_0))$, then the solution is unique.

Proof. let $e$ be a $(p,\alpha)$ Lebesgue number. Then, $M = \bigcup_{i=1}^N S_i$ where $\operatorname{diam}(S_i) < e$. Moreover allow $m_0 \in S_1$ ($S_i$ need not be closed or open or anything, but it should be connected).

Now, define $M_t = \bigcup_{i=1}^t S_i$, $\alpha_t = \alpha|_{M_t}$. we say that the lifting problem has a $t$-solution if there exists a unique $\tilde{\alpha}_t$ such that $\tilde{\alpha}_t(m_0) = x_0$, and (of course) $p \circ \tilde{\alpha}_t = \alpha_t$.

We shall proceed inductively, solving instances using the relative lifting lemma. As usual, we show the inductive step first.

Suppose that this lifting problem has a $t$-solution. Denote $\alpha|_{S_{t+1}}$ as $\alpha^*_{t+1}$, $\tilde{\alpha}_t|_{M_t \cap S_{t+1}}$ as $\beta_t$, mainly for convenience.

Since each $S_i$ is connected, and $M_t \cap S_{t+1}$ is a connected subset of $S_{t+1}$, and of course, $\alpha(S_i) \subset U$ where $U$ is evenly covered, the local lifting lemma says, there exists a unique $\tilde{\alpha}^*_{t+1}$ for which, $p \circ \tilde{\alpha}^*_{t+1} = \alpha^*_{t+1}$ and $\tilde{\alpha}^*_{t+1}(M \cap S_{t+1}) = \beta_t(M \cap S_{t+1}) = \tilde{\alpha}_t(M \cap S_{t+1})$.

Pause here for a second, it is no coincidence that we choose the lift relative to the parts where $M_t$ and $S_{t+1}$ overlap! we will glue these solutions together!!

Define

$$\tilde{\alpha}_{t+1}\colon M_{t+1}=M_t\cup S_{t+1}\longrightarrow X, \qquad \tilde{\alpha}_{t+1}(m)= \begin{cases} \tilde{\alpha}_t(m), & m\in M_t,\\[4pt] \tilde{\alpha}^{*}_{t+1}(m), & m\in S_{t+1}. \end{cases}$$

This map is unique, since it is the piecewise join of the unique maps $\tilde{\alpha}_t$ and $\tilde{\alpha}^{*}_{t+1}$. The two maps agree on $M_t\cap S_{t+1}$, so the gluing lemma makes $\tilde{\alpha}_{t+1}$ continuous. Finally, since $m_0\in M_t$, we still have $\tilde{\alpha}_{t+1}(m_0)=\tilde{\alpha}_t(m_0)=x_0$.

The base case is done by applying the local lifting lemma to $S_1$, and doing it relative to $\{m_0\}$, which is a single-element connected subset of $S_1$.

2 - Lifting Paths and Homotopies

Path Lifting

Awesome! so now we can lift paths! can we lift loops? Not really, as $p$ need not be injective.

Homotopy Lifting

We can also lift homotopies! set $M = [0,1]^2$ if the homotopy $H\colon [0,1]^2 \to Y$ fixes endpoints, that is, $H(0,\_) = c_0$ and $H(1, \_) = c_1$ (where $c_0, c_1$ are constant functions) meaning that at any homotopy slice $H$ has the same two endpoints.

The left interval and right interval are of course connected, so their images under $\tilde{H}$ must also be connected, but the preimage of $y_0$ under $p$ is a bunch of disconnected points, one on each sheet, thus, $\tilde{H}$ is forced to pick one of those points. i.e. $\tilde{H}$ also fixes endpoints.

In particular, this means that though lifts of homotopy paths with fixed endpoints have fixed endpoints, the lift of a homotopy of loops, may not be a homotopy of loops but just a homotopy of paths.

3 - The Exponential Covering Map

Now, we go back to the map we’re dealing with.

where $p(x) = e^{i2 \pi x}$, we shall show $p$ is a covering map.

for convenience, we will take the argument of a complex number to lie in $[0,2\pi)$, where indeed, $z = e^{ia(z)}$ where $a(z)$ is this modified argument.

An open set in $S^1$ can be given by an open interval $I \subset [0, 2\pi)$ as $A = \{e^{it} \mid t \in I\}$. for each $z \in S^1$ choose $1/2 > e > 0$ such that $I_z =(a(z) - e, a(z) + e) \subset [0,2\pi)$. we claim that $E_z = \{e^{it} \mid t \in I_z\}$ is evenly covered by $p$.

Notice,

$p^{-1}(E_z) = \bigsqcup_{k \in \mathbb{Z}} \left(\frac{a(z)-e}{2\pi} + k, \frac{a(z) + e}{2\pi} + k\right).$

call such an interval $I_z^k = \left(\frac{a(z)-e}{2\pi} + k, \frac{a(z) + e}{2\pi} + k\right)$. We choose $e < 1/2$ as then the length of this interval is $2e/2\pi$ which is well smaller than $1$.

we claim that for any $k \in \mathbb{Z}$, $p^* \coloneqq p|_{I^k_z}\colon I^k_z \to E_z$ is a homeomorphism.

of course, $p^*$ is continuous, as it is a restriction of a continuous map. now, $p^*(I^k_z) = \{e^{i2\pi x} \mid x \in I^k_z\}$.

Notice however that $2\pi x \in \left({a(z)-e} + 2\pi k, a(z) + e + {2\pi}k\right)$ but obviously, $e^{i2\pi x} = e^{i(2\pi x - 2\pi k)}$.

$2\pi x - 2\pi k \in \left({a(z)-e}, a(z) + e \right) = I_z$ Therefore $p^{*}(I^k_z) = \{e^{it} \mid t \in I_z\} = E_z$.

so $p^*$ is surjective. now, suppose $p^*(x) = p^*(y)$, then $e^{i2\pi x} = e^{i 2\pi y}$. without loss of generality $y \ge x$, then we get $y = x + h$ for some positive integer $h$. of course, since $y,x \in I^k_z$, $y-x < 1$, thus $y = x$. Indeed $p^*$ is injective.

Define

$$q_k\colon E_z\longrightarrow I_z^k, \qquad q_k(v)=\frac{a(v)}{2\pi}+k.$$

Here $a\colon E_z\to I_z$ is the continuous branch of the argument determined by $v=e^{ia(v)}$. We claim that $q_k=(p^*)^{-1}$.

First, if $v\in E_z$, then

$$(p^*\circ q_k)(v) =p^*\left(\frac{a(v)}{2\pi}+k\right) =e^{i2\pi\left(\frac{a(v)}{2\pi}+k\right)} =e^{ia(v)}e^{i2\pi k} =v.$$

Thus $p^*\circ q_k=\operatorname{id}_{E_z}$.

Conversely, let $x\in I_z^k$. Then $x-k\in I_z^0$, so $2\pi(x-k)\in I_z$. Since $a$ takes values in $I_z$, it follows that

$$a(p^*(x))=a(e^{i2\pi x})=2\pi(x-k).$$

Therefore

$$(q_k\circ p^*)(x) =\frac{a(p^*(x))}{2\pi}+k =\frac{2\pi(x-k)}{2\pi}+k =x.$$

Hence $q_k\circ p^*=\operatorname{id}_{I_z^k}$ as well, proving that $q_k=(p^*)^{-1}$.

Finally, $q_k$ is continuous because it is obtained from the continuous local argument $a|_{E_z}$ by scaling by $1/(2\pi)$ and translating by $k$. Thus $p^*\colon I_z^k\to E_z$ is a homeomorphism. Since this holds for every $k\in\mathbb Z$, the set $E_z$ is evenly covered by $p$.

Lifting Loops

To proceed, let us emphasize what the compact-metric lifting theorem says. it says, continuous maps from a compact metric space to a topological space can be lifted along a covering map to another topological space. In particular, it also says this lift is unique as long as its restriction on a connected subset of the metric space is determined.

choose $k \in \mathbb Z$. Then, consider the set of loops on $S^1$ with basepoint $\mathbf{1} \in \mathbb{C}$, denoted by $\Omega(S^1, \mathbf{1})$, and consider the set of paths $P(\mathbb R)$ on the reals. The compact-metric lifting theorem says that the map $L_k\colon \Omega(S^1,\mathbf{1}) \to P(\mathbb R)$ given by $\alpha \mapsto \tilde{\alpha}_k$ where $\tilde{\alpha}_k(0) = k$ is well defined, and is injective. ($\{0\}$ is a connected subset of $[0,1]$)

That is $L_k$ takes a loop, and gives the unique lift of it, for which its image of zero is $k$.

at this point, I thought that any real number shift of a lift is another lift, I quickly realized that of course, this is not true, if you lift a loop, shift it only a tiny amount and project it back down, it is still morally the same loop but the basepoint has shifted, however, an integer shift will wrap back around to the same basepoint

Shifting Lifts

we claim that for any integer $r$, $r \oplus \tilde{\alpha}_k\colon x \mapsto r+ \tilde{\alpha}_k(x)$ is a lift of $\alpha$.

obviously, $p(r + \tilde{\alpha}_k(x)) = e^{i2\pi(r + \tilde{\alpha}_k(x))} = p(\tilde{\alpha}_k(x)) = \alpha(x)$

in particular, $(r \oplus \tilde{\alpha}_k)(0) = k + r$, but $\tilde{\alpha}_{k+r}$ is the unique lift whose image of zero is $k + r$. Therefore, $r \oplus \tilde{\alpha}_k = \tilde{\alpha}_{k+r}$ whenever $r$ is an integer.

The Endpoint Map

Another map we can define is $\phi_k\colon \Omega(S^1, \mathbf{1}) \to \mathbb{Z}$ given by $\alpha \mapsto \tilde{\alpha}_k(1) - \tilde{\alpha}_k(0)$. Well, why is the difference an integer? $\alpha$ is a loop, and $\alpha(1) = \alpha(0) = \mathbf{1}$. thus, $\tilde{\alpha}_k(1), \tilde{\alpha}_k(0) \in p^{-1}(\mathbf{1}) = \mathbb{Z}$.

Homotopy Invariance

Now, we claim that whenever $\alpha \sim \beta$ by homotopy of loops, then $\phi_k(\alpha) = \phi_k(\beta)$.

let $H\colon [0,1]^2 \to S^1$ be a homotopy from $\alpha$ to $\beta$.

of course, we can lift $H$, choose the unique lift $\tilde{H}$ for which $\tilde{H}(0,0) = k$.

of course, as shown earlier, if $H$ fixes endpoints, then so does $\tilde{H}$. That is to say, $\tilde{H}(0,\_)$ and $\tilde{H}(1,\_)$ are constant functions, moreover this means, $\tilde{H}(0,\_) =k$ as a constant function.

obviously, $\tilde{H}(\_,0), \tilde{H}(\_,1)$ are lifts of $H(_,0) = \alpha, H(_,1) = \beta$ respectively, moreover, they are the unique lifts whose image at $0$ is $k$.

Indeed, $\tilde{H}(_,0) = \tilde{\alpha}_k, \ \tilde{H}(_, 1) =\tilde{\beta}_k$

well, of course then the endpoints of $\tilde{\alpha}_k, \tilde{\beta}_k$ are respectively equal, i.e. $\phi_k(\alpha) = \phi_k(\beta)$.

Thus, we can define without worry $\Phi_k\colon \pi_1(S^1,\mathbf{1}) \to \mathbb{Z}$ by $[\alpha] \mapsto \phi_k(\alpha)$.

Wow, this is a wild amount of setup!

The Fundamental Group of the Circle

Well, we may as well inspect $\Phi_0$, given by $[\alpha] \mapsto \tilde{\alpha}_0(1)$. We will show that it is a group homomorphism. i.e. $\Phi_0([\alpha * \beta]) = \tilde{\alpha}_0(1) + \tilde{\beta}_0(1)$.

We can tell that we want to lift $\alpha * \beta$ into some path $p$ such that $p(0) = 0$ and $p(1) = \tilde{\alpha}_0(1) + \tilde{\beta}_0(1)$.

Naively, one might think of joining $\tilde{\alpha}_0$ with $\tilde{\beta}_0$, but their only common point might be $0$ itself, in particular, it may be that $\tilde{\alpha}_0(1) \neq 0 = \tilde{\beta}_0(0)$. But what we do know, is that $\tilde{\alpha}_0(1) = q \in \mathbb Z$!

As seen earlier, $\tilde{\beta}_q = q \oplus \tilde{\beta}_0$.

we claim that the lift with startpoint $0$ of the product is

$$L_0(\alpha * \beta)(t) = \begin{cases} \tilde{\alpha}_0(2t) & t \in [0,1/2], \\ \tilde{\beta}_q(2t-1) & t \in [1/2,1]. \end{cases}$$

composing the above map with the covering map $p$ piecewise, it is not hard to see that it lifts $\alpha * \beta$. of course, the map agrees at the common point, (notice $q = \tilde{\alpha}_0(1)$). and $(L_0(\alpha * \beta))(0) = 0$. (as defined)

Now notice that $L_0(\alpha * \beta)(1) = \tilde{\beta}_q(1) = q + \tilde{\beta}_0(1) = \tilde{\alpha}_0(1) + \tilde{\beta}_0(1) = \Phi_0([\alpha * \beta])$.

i.e. $\Phi_0$ is a homomorphism of groups!

we shall next show that $\Phi_0$ is injective. suppose $\Phi_0([\alpha]) = \Phi_0([\beta])$ then of course $\tilde{\alpha}_0(1) = \tilde{\beta}_0(1)$, i.e. $\tilde{\alpha}_0, \tilde{\beta}_0$ have the same endpoints. (at this point, we think these two paths are homotopic, and we can project it down via $p$ to show $\alpha$ is homotopic to $\beta$)

Define $U\colon [0,1]^2 \to \mathbb{R}$ given by $U(t,s) = (1-s)\tilde{\alpha}_0(t) + s \tilde{\beta}_0(t)$. Indeed, $U$ is a homotopy (with endpoints fixed) from $\tilde{\alpha}_0$ to $\tilde {\beta}_0$. (The reader is asked to pause and verify.)

so, $p \circ U\colon [0,1]^2 \to S^1$ is indeed a continuous map. Since both endpoints, namely $0$, $\tilde{\alpha}_0(1) = \tilde{\beta}_0(1) \in \mathbb Z$, $p \circ U(0, \_) = p(0) = \mathbf{1}$ and $p \circ U(1, \_) = p(\tilde{\alpha}_0(1)) = \mathbf{1}$, we have each slice $p \circ U(\_,s)$ is a loop in $S^1$.

of course, $p \circ U(_,0) = p \circ \tilde{\alpha}_0 = \alpha$ and similarly, $p \circ U(_, 1) = \beta$

thus $p \circ U$ is a homotopy of loops, from $\alpha$ to $\beta$, i.e. $[\alpha] = [\beta]$.

A nice thing about showing $\Phi_0$ is a homomorphism, is that it is enough to show a loop $\alpha$ for which $\Phi_0([\alpha]) = 1 \in \mathbb{Z}$ to show surjectivity.

consider the loop $p|_{[0,1]}\colon [0,1] \to S^{1}$ it is clear that the inclusion map $\iota\colon [0,1] \to \mathbb R$ lifts this loop, (by definition, $p|_{[0,1]} = p \circ \iota$). and of course $\Phi_0([p|_{[0,1]}]) = \iota(1) = 1$.

It directly follows that $\Phi_0([(p|_{[0,1]})^{-1}]) = -1$. And thus, any integer is formed by taking $\Phi_0$ on repeated products.

Thus finally $\Phi_0\colon \pi_1(S^1, \mathbf{1}) \to \mathbb{Z}$ is a bijective homomorphism. i.e.

$\pi_1(S^1,1) \cong \mathbb Z$

LET’S GO!!

4 - Homotopy Equivalence and Fundamental Groups

Recall that two topological spaces $X,Y$ (with base points) are homotopy equivalent if there exists a pair of continuous basepoint preserving maps $f\colon X \to Y, g\colon Y \to X$ such that $g \circ f$ is homotopic to $\operatorname{id}_X$ and $f \circ g$ is homotopic to $\operatorname{id}_Y$.

if the fundamental group functor $\pi_1\colon \mathsf{Top}_* \to \mathsf{Grp}$ is really about homotopy, then homotopy equivalent spaces should have isomorphic fundamental groups.

The above diagram isn’t exactly a commutative diagram in some fixed category, but it uses the fact that $\pi_1$ takes continuous maps to a new map, that sends a homotopy class of loops to the class induced by composing a representative with the map. i.e. $\pi_1(f)\colon [\alpha] \mapsto [f \circ \alpha]$.

due to functoriality, $\pi_1(g) \circ \pi_1(f) = \pi_1(g\circ f)\colon [\alpha] \mapsto [g \circ f \circ \alpha]$.

let $H\colon X \times [0,1] \to X$ be a homotopy from $g\circ f$ to $\operatorname{id}_X$.

$H \circ (\alpha \times \operatorname{id}_{[0,1]})\colon [0,1]^2 \to X$ is a homotopy from $g \circ f \circ \alpha$ to $\alpha$, i.e. $g\circ f \circ \alpha \sim \alpha$. similarly if $\beta$ is a loop on $Y$, then $f\circ g \circ \beta \sim \beta$.

i.e. $\pi_1(g), \ \pi_1(f)$ are inverses of each other.

i.e. $\pi_1(X) \cong \pi_1(Y).$

The Punctured Plane and the Circle

Now, finally we will show that $\mathbb C - \{0\} = \mathbb C_*$ is homotopy equivalent to $S^1$.

set $f\colon \mathbb C_* \to S^1$ defined by $z \mapsto z/|z|$ and $\iota\colon S^1 \to \mathbb C_*$ the inclusion map. of course, $f \circ \iota = \operatorname{id}_{S^1}$ as is, now, consider $\iota \circ f\colon \mathbb C_* \to \mathbb C_*$.

we want to construct a homotopy $H\colon \mathbb C_* \times [0,1] \to \mathbb C_*$ that is continuous, and takes $\operatorname{id}_{\mathbb C_*}$ to $\iota \circ f$.

as usual we will fade one function out, and fade the other one in.

$H(z,t) = (1-t)z + \frac{tz}{|z|} = \left((1-t) + \frac{t}{|z|}\right)z$

Notice carefully that we are quite illiterate in analysis, so we will prove the continuity of the maps involved in a later note (it should be pretty chill, however this note is already getting too long).

But it does seem like continuous functions into the complex numbers can be added, composed and multiplied and if you removed the zero, can also be inverted. This is an interesting aside,

5 - Continuous Maps into Topological Fields

Enter Topological Fields.

A topological space $(X,+,\cdot)$ is a topological field, if it’s a field such that the following maps are continuous.

$+,\cdot\colon X^2 \to X$ (product topology on $X^2$) and $({\_})^{-1}\colon X_* \to X_*$ ($X$ without zero).

Let $X$ be a topological field and $Y$ a topological space. Let

$$C(Y,X)=\{f\colon Y\to X\mid f\text{ is continuous}\}.$$

Define the operations pointwise.

$$(f+g)(y)=f(y)+g(y),$$ $$(fg)(y)=f(y)g(y).$$

And if $f(y)\neq 0$ for every $y$, define

$$f^{-1}(y)=f(y)^{-1}.$$

Now we just need to check that these things are actually continuous.

If $f,g\in C(Y,X)$, then

$$(f,g)\colon Y\to X^2,\qquad y\mapsto(f(y),g(y))$$

is continuous, since $f$ and $g$ are continuous and $X^2$ has the product topology.

Warning. The above diagram is NOT A COMMUTATIVE DIAGRAM, I have deliberately drawn it that way, it is just to illustrate the compositions on each path.

Indeed, if you follow each path pointwise, you get $fg\colon Y \to X$, $f+g\colon Y \to X$, $f^{-1}\colon Y \to X_*$, which are compositions of continuous maps.

That is to say, continuous maps into topological fields are closed with respect to algebraic operations, (the inverse is the pesky one as usual).

Note that we are not saying that these maps themselves form a field (they form a ring of course).

Now we need to show that $\mathbb C$ is a topological field, we will do so in a later note.

6 - No Continuous Square Root

Now, back to the main thread.

The machinery we have built so far is powerful, and will be used over and over. Let us do a demonstration.

We claim that there is no continuous square-root function on $\mathbb C$. Write

$$s\colon\mathbb C\longrightarrow\mathbb C, \qquad s(z)=z^2.$$

A continuous square root would be a map $r\colon\mathbb C\to\mathbb C$ such that $s\circ r=\operatorname{id}_{\mathbb C}$; that is, $r$ would be a section of the squaring map.

if an $r$ existed on the non-zero complex numbers, we could bring it back to all of $\mathbb C$ by filling in $r(0)=0$.

indeed, take any open ball $B(0,e)$ around the origin in the codomain. since $r(z)^2=z$, we have

$$r^{-1}(B(0,e))=B(0,e^2).$$

the right hand side is an open ball around the origin in the domain. Thus the extension is continuous at $0$, and it was already continuous everywhere else.

Therefore, it is enough to show that there is no continuous square root on $\mathbb C_*=\mathbb C\setminus\{0\}$.

Assume for contradiction that $r\colon\mathbb C_*\to\mathbb C_*$ exists.

Since $r(1)=\pm1$, replacing $r$ by the continuous square root $(-r)(z)=-r(z)$ if necessary lets us assume $r(1)=1$. Thus $r$ and $s$ are based maps at $1$, and $s\circ r=\operatorname{id}_{\mathbb C_*}$.

Applying $\pi_1$ gives

$$\pi_1(s)\circ\pi_1(r) =\pi_1(s\circ r) =\operatorname{id}_{\pi_1(\mathbb C_*,1)}.$$

The Fundamental Group Obstruction

We claim that the map induced by squaring is the doubling map.

$$\pi_1(s)=d\colon\mathbb Z\longrightarrow\mathbb Z, \qquad d(n)=2n.$$

Assuming this claim, call $h=\pi_1(r)$. The equation above becomes

$$d\circ h=\operatorname{id}_{\mathbb Z}.$$

But no such $h$ can exist. plugging in $1$ would give $2h(1)=1$, which is impossible since $h(1)$ is an integer.

Squaring Induces Doubling

Proof of the claim. first, let us be clear about what the element $1\in\mathbb Z$ represents. Recall that

$$\Phi_0\colon\pi_1(S^1,\mathbf 1)\longrightarrow\mathbb Z$$

sends the class of the loop

$$\gamma=p|_{[0,1]},\qquad \gamma(t)=e^{i2\pi t},$$

to $1$, since its lift starting at $0$ is $t\mapsto t$ and ends at $1$.

Now let $\iota\colon S^1\hookrightarrow\mathbb C_*$ be the inclusion. We have already shown that

$$\iota_*\colon\pi_1(S^1,\mathbf 1)\longrightarrow\pi_1(\mathbb C_*,\mathbf 1)$$

is an isomorphism. Thus, when we identify $\pi_1(\mathbb C_*,\mathbf 1)$ with $\mathbb Z$ using $\Phi_0\circ\iota_*^{-1}$, we get

$$(\Phi_0\circ\iota_*^{-1})(\iota_*([\gamma])) =\Phi_0([\gamma])=1.$$

So the same loop, now regarded as a loop in $\mathbb C_*$, represents the element $1$ there as well.

well, $\pi_1(s)\colon\mathbb Z\to\mathbb Z$ is a group homomorphism, and a group homomorphism out of $\mathbb Z$ is completely determined by where it sends $1$.

then

$$(s\circ\gamma)(t)=\gamma(t)^2=e^{i4\pi t}.$$

the lift of this loop which starts at $0$ is just $t\mapsto 2t$, and it ends at $2$. Thus $\pi_1(s)$ sends $1$ to $2$.

but now the entire homomorphism is forced. for every $n\in\mathbb Z$,

$$\pi_1(s)(n)=n\pi_1(s)(1)=2n.$$

Therefore $\pi_1(s)=d$, as claimed.