Null Homotopic Maps: RP3 ∨ RP3 To S1 ∨ S1

Hey guys! Today, we're diving deep into the fascinating world of algebraic topology, specifically looking at maps between topological spaces and trying to figure out if they're null homotopic. Null-homotopic maps are basically maps that can be continuously deformed into a constant map (a map that sends everything to a single point). We are going to show any map from $\Bbb{R}\Bbb{P}^3 \vee \Bbb{R}\Bbb{P}^3$ to $S^1 \vee S^1$ is null homotopic. Buckle up, because this is going to be a wild ride!

Understanding the Spaces: Real Projective Space and the Circle

Before we jump into the main problem, let's make sure we're all on the same page about the spaces involved. We have two main characters here: real projective space ($\mathbb{RP}^3$) and the circle ($S^1$).

Real Projective Space: $\mathbb{RP}^3$

Real projective space, denoted as $\mathbb{RP}^n$, can be thought of as the set of all lines through the origin in $\mathbb{R}^{n+1}$. In other words, each point in $\mathbb{RP}^n$ corresponds to a line in $\mathbb{R}^{n+1}$. A common way to visualize $\mathbb{RP}^n$ is by identifying antipodal points on the $n$-sphere $S^n$. For example, $\mathbb{RP}^1$ is homeomorphic to the circle $S^1$, and $\mathbb{RP}^2$ is the projective plane. For our problem, we're specifically interested in $\mathbb{RP}^3$.

$\mathbb{RP}^3$ is a three-dimensional manifold that can be constructed by taking the 3-sphere $S^3$ and identifying antipodal points. This means that for every point $(x, y, z, w)$ on $S^3$, we identify it with the point $(-x, -y, -z, -w)$. This identification process gives $\mathbb{RP}^3$ some interesting topological properties. One key property is its fundamental group.

The fundamental group of $\mathbb{RP}^3$, denoted as $\pi_1(\mathbb{RP}^3)$, is isomorphic to $\mathbb{Z}_2$, the cyclic group of order 2. This means that there are two homotopy classes of loops in $\mathbb{RP}^3$: the trivial loop (which can be contracted to a point) and a non-trivial loop that represents the generator of $\mathbb{Z}_2$. This non-trivial loop corresponds to a path that connects a point on $S^3$ to its antipodal point.

The Circle: $S^1$

The circle, denoted as $S^1$, is simply the set of all points in the plane that are a fixed distance (usually 1) from the origin. Topologically, it's a one-dimensional manifold that's probably the simplest non-trivial space you can imagine. Its fundamental group, $\pi_1(S^1)$, is isomorphic to $\mathbb{Z}$, the infinite cyclic group. This means that there are infinitely many homotopy classes of loops in $S^1$, each corresponding to the number of times the loop winds around the circle (positive or negative).

The Wedge Sum: $X \vee X$ and $Y \vee Y$

Now that we understand $\mathbb{RP}^3$ and $S^1$, let's talk about the wedge sum operation. The wedge sum of two topological spaces, denoted as $X \vee Y$, is formed by taking the disjoint union of $X$ and $Y$ and then identifying a single point in $X$ with a single point in $Y$. Think of it as gluing the two spaces together at a single point.

In our case, we have $\mathbb{RP}^3 \vee \mathbb{RP}^3$ and $S^1 \vee S^1$. The first one is two copies of $\mathbb{RP}^3$ glued together at a single point, and the second one is two circles glued together at a single point (which looks like a figure eight!).

The fundamental group of a wedge sum is the free product of the fundamental groups of the individual spaces. That is, $\pi_1(X \vee Y) = \pi_1(X) * \pi_1(Y)$. Therefore, we have:

• $\pi_1(\mathbb{RP}^3 \vee \mathbb{RP}^3) = \pi_1(\mathbb{RP}^3) * \pi_1(\mathbb{RP}^3) = \mathbb{Z}_2 * \mathbb{Z}_2$
• $\pi_1(S^1 \vee S^1) = \pi_1(S^1) * \pi_1(S^1) = \mathbb{Z} * \mathbb{Z}$

The group $\mathbb{Z}_2 * \mathbb{Z}_2$ is the free product of two cyclic groups of order 2, and $\mathbb{Z} * \mathbb{Z}$ is the free group on two generators.

The Problem: Showing the Map is Null Homotopic

Okay, with all the background out of the way, let's finally get to the heart of the problem. We want to show that any map $f: \mathbb{RP}^3 \vee \mathbb{RP}^3 \to S^1 \vee S^1$ is null homotopic. In other words, we want to show that $f$ can be continuously deformed into a constant map.

To tackle this, we're going to use the fundamental group. Recall that a map $f: X \to Y$ induces a homomorphism $f_*: \pi_1(X) \to \pi_1(Y)$ between the fundamental groups of $X$ and $Y$. If $f$ is null homotopic, then $f_*$ must be the trivial homomorphism (i.e., it sends every element of $\pi_1(X)$ to the identity element in $\pi_1(Y)$).

So, our strategy is to show that any homomorphism from $\pi_1(\mathbb{RP}^3 \vee \mathbb{RP}^3)$ to $\pi_1(S^1 \vee S^1)$ must be the trivial homomorphism. This will imply that any map $f: \mathbb{RP}^3 \vee \mathbb{RP}^3 \to S^1 \vee S^1$ is null homotopic.

The Proof

Let $f: \mathbb{RP}^3 \vee \mathbb{RP}^3 \to S^1 \vee S^1$ be any map. This map induces a homomorphism

$$f_*: \pi_1(\mathbb{RP}^3 \vee \mathbb{RP}^3) \to \pi_1(S^1 \vee S^1)$$

which is equivalent to

$$f_*: \mathbb{Z}_2 * \mathbb{Z}_2 \to \mathbb{Z} * \mathbb{Z}$$

Now, let's consider the generators of $\mathbb{Z}_2 * \mathbb{Z}_2$. Let $a$ and $b$ be the generators of the two $\mathbb{Z}_2$ groups. This means that $a^2 = e$ and $b^2 = e$, where $e$ is the identity element. So, $a$ and $b$ are elements of order 2.

If $f_*$ is any homomorphism, then the order of $f_*(a)$ must divide the order of $a$, and the order of $f_*(b)$ must divide the order of $b$. In other words, the order of $f_*(a)$ and $f_*(b)$ must divide 2. This means that $f_*(a)^2 = e'$ and $f_*(b)^2 = e'$, where $e'$ is the identity element in $\mathbb{Z} * \mathbb{Z}$.

However, the only element in $\mathbb{Z} * \mathbb{Z}$ with finite order is the identity element. This is because $\mathbb{Z} * \mathbb{Z}$ is a free group, and free groups have no elements of finite order (except for the identity).

Therefore, we must have $f_*(a) = e'$ and $f_*(b) = e'$. This means that $f_*$ sends both generators of $\mathbb{Z}_2 * \mathbb{Z}_2$ to the identity element in $\mathbb{Z} * \mathbb{Z}$. Since $a$ and $b$ generate $\mathbb{Z}_2 * \mathbb{Z}_2$, it follows that $f_*$ is the trivial homomorphism.

Since $f_*$ is the trivial homomorphism, the map $f: \mathbb{RP}^3 \vee \mathbb{RP}^3 \to S^1 \vee S^1$ is null homotopic. And that's what we wanted to show!

Conclusion

So, there you have it! We've successfully shown that any map from $\mathbb{RP}^3 \vee \mathbb{RP}^3$ to $S^1 \vee S^1$ is null homotopic. This result relies on understanding the fundamental groups of these spaces and the properties of homomorphisms between them. It's a cool example of how algebraic topology can be used to study the properties of topological spaces and maps between them.

Hope you guys enjoyed this deep dive into the world of algebraic topology. Keep exploring, and keep asking questions!