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 $F_{*}$?
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.