Commutative Diagrams and their relationship to induced homomorphisms

Question

Throughout my topology class my professor has used commutative diagrams on various occasions to prove results such as

1) There exists no antipode preserving, continuous, onto map, $f: S^2 \to S^1$

2) $[0,1] \setminus 0 \sim 1$ is homeomorphic to $S^1$.

3) Prove that $S^2 \setminus (x,y,z) \sim (-x,-y,-z)$ is homeomorphic to the real projective plane.

The diagram of (1) looks like this,

$$\begin{array} ^{S^2} & \stackrel{f}{\longrightarrow} & S^1 \\ \downarrow{q_1} & & \downarrow{q_2} \\ S^2 \setminus \sim & \stackrel{F}{\longrightarrow} & S^1\setminus \sim \end{array} $$

where $S^2 \setminus \sim$ refers to $S^2 \setminus (x,y,z) \sim (-x,-y,-z)$ and $S^1 \setminus x \sim -x$

$f:S^2 \to S^1$ is such that $f(x)=-f(-x)$. To prove such a function does not exists we used the commutative diagram of the induced homomorphism.

$$\begin{array} ^\pi_1({S^2}) & \stackrel{f_*}{\longrightarrow} & \pi_1(S^1) \\ \downarrow{q_{1_*}} & & \downarrow{q_{2_*}} \\ S^2 \setminus \sim & \stackrel{F_*}{\longrightarrow} & S^1\setminus \sim \end{array} $$

and showed that an appropriate $F*$ does not exists.

$\textbf{Question}$ Why does the non-existense of $F_{*}$ imply that $f$ cannot exist?

Furthermore, what is meant by an appropriate $F_{*}$?

Answer 1

The proposition we wish to prove is that there does not exist a map $f\colon S^2\to S^1$ such that $f(x)=-f(-x)$ for all $x\in S^2$. In order to prove this, let us suppose that such an $f$ does exists. Given this, there must then also exist an $F\colon S^2/{\sim}\to S^1/{\sim}$ given by $F([x])=[f(x)]$. In order to see that $F$ is well defined we note that if $x'\in[x]$ then $x'$ is either $x$ or $-x$ by definition of $\sim$ and so the only other possible value of $F([-x])$ is $[f(-x)]$ but by the condition that $f$ satisfies, we know that $f(-x)=-f(x)$ and so $F([-x])=[-f(x)]$ and again by the definition of $\sim$ (the one on $S^1$ this time) we know that $[-f(x)]=[f(x)]$. It follows that $F([-x])=[f(x)]=F([x])$ and so $F$ is well defined.

The fact that $F$ is continuous is a consequence of the universal property of quotient maps and the fact that we have the commutative diagram $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} S^2 & \ra{f} & S^1 \\ \da{q_1} & & \da{q_2} \\ S^2/{\sim} & \ras{F} & S^1/{\sim} \\ \end{array} $$ where $q_i$ is the appropriate quotient map.

So, given that $f$ exists with the required property, there must exist an $F$ which satisfies the above commutative square.

By applying the fundamental group functor to the above diagram of continuous maps, we get the following diagram

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} \pi_1(S^2) & \ra{f_*} & \pi_1(S^1) \\ \da{(q_1)_*} & & \da{(q_2)_*} \\ \pi_1(S^2/{\sim}) & \ras{F_*} & \pi_1(S^1/{\sim}) \\ \end{array} $$

and we note that $\pi_1(S^2/{\sim})\cong\mathbb{Z}/2\mathbb{Z}$. It follows that $F_*$ is the zero map as this is the only homomorphism from a finite group into the integers. This means that no loop in $S^2/{\sim}$ is mapped to a non-trivial loop in $S^1$ under $F$. By considering the equator $A\cong S^1$ of $S^2$, we see that $(f|_A)_*\colon\pi_1(A)\to \pi_1(S^1)$ is not the zero map as $f|_A$ is antipode preserving, and $(q_2)_*$ is just the multiplication by $2$ map and so there exists some loop $\gamma$ living in $A\subset S^2$ which is mapped to a non-trivial loop in $S^1/{\sim}$. This means, by commutativity, that $q_1\circ\gamma$ is a loop in $S^2/{\sim}$ which is mapped to a non-trivial loop in $S^1/{\sim}$ - however this contradicts the fact that we said no such loop can exist.

It follows that we have made a false assumption somewhere. The only assumption made was that $f$ with the required properties exists. It follows such an $f$ can not exist and so we are done.