Answer: The Wedge That Knows Its Loops

Key Idea / Intuition

The fundamental group of a wedge is the free product of the fundamental groups (by van Kampen's theorem). A connected $n$-sheeted covering space of $X$ corresponds to a subgroup of index $n$ in $\pi_1(X)$. A 2-sheeted covering exists if and only if $\pi_1(X)$ has a subgroup of index 2 — equivalently, a surjective homomorphism $\pi_1(X) \to \mathbb{Z}/2\mathbb{Z}$. The free product structure makes this easy to detect, and the answer reveals a beautiful interplay between torsion and the abelianization.

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Formal Proof / Solution

Step 1: Compute $\pi_1(X)$.

By the Seifert–van Kampen theorem, the fundamental group of a wedge $A \vee B$ (with the basepoint at the wedge point) is the free product: $\pi_1(A \vee B) \cong \pi_1(A) * \pi_1(B).$

We know:

• $\pi_1(\mathbb{R}P^2) \cong \mathbb{Z}/2\mathbb{Z}$, generated by $a$ with $a^2 = 1$.
• $\pi_1(S^1) \cong \mathbb{Z}$, generated by $b$.

Therefore: $\pi_1(X) \cong \mathbb{Z}/2\mathbb{Z} * \mathbb{Z} = \langle a, b \mid a^2 = 1 \rangle.$

Step 2: Does a 2-sheeted covering exist?

A connected 2-sheeted covering of $X$ corresponds exactly to a normal subgroup of index 2 in $\pi_1(X)$, or equivalently, a surjective homomorphism $\phi: \pi_1(X) \to \mathbb{Z}/2\mathbb{Z}$.

Since $\pi_1(X) = \langle a, b \mid a^2 = 1 \rangle$, a homomorphism $\phi$ to $\mathbb{Z}/2\mathbb{Z} = \{0,1\}$ is determined by the images of $a$ and $b$, subject only to $\phi(a)^2 = 0$. In $\mathbb{Z}/2\mathbb{Z}$ every element satisfies $x^2 = 0$, so any assignment of $\phi(a), \phi(b) \in \{0,1\}$ works.

For surjectivity, at least one generator must map to $1$. One natural choice: $\phi(a) = 1, \quad \phi(b) = 0.$

So yes, a connected 2-sheeted covering does exist.

Step 3: Describe the covering space.

The covering corresponding to $\ker \phi$ (with $\phi(a)=1, \phi(b)=0$) can be described geometrically. The covering of $\mathbb{R}P^2$ corresponding to the unique index-2 subgroup $\{e\} \subset \mathbb{Z}/2\mathbb{Z}$ is $S^2$ (the universal cover of $\mathbb{R}P^2$). The covering of $S^1$ corresponding to the whole group $\mathbb{Z}$ (index-2 subgroup $2\mathbb{Z}$ ignored since $\phi(b)=0$) is two disjoint copies of $S^1$.

More precisely, the 2-sheeted covering $\tilde{X}$ is: $\tilde{X} \cong S^2 \vee S^1 \vee S^1.$

Why? The deck transformation swaps the two sheets. Over the $\mathbb{R}P^2$ factor (which $\phi$ "sees"), the covering lifts to $S^2$. Over the $S^1$ factor (which $\phi$ "doesn't see," since $\phi(b)=0$), the preimage consists of two copies of $S^1$, each living in one sheet, both attached at the lifted basepoint. The resulting space is $S^2$ with two circles attached at a common point, i.e., $S^2 \vee S^1 \vee S^1$.

Sanity check via $\pi_1$: $\pi_1(S^2 \vee S^1 \vee S^1) \cong \mathbb{Z} * \mathbb{Z},$ which is indeed the index-2 subgroup $\ker\phi = \langle b, aba^{-1} \rangle \cong \mathbb{Z}*\mathbb{Z}$ in $\mathbb{Z}/2\mathbb{Z} * \mathbb{Z}$ (the subgroup generated by $b$ and the conjugate $aba^{-1} = aba$, since $a^{-1}=a$).

Summary:

| | Value | |---|---| | $\pi_1(X)$ | $\mathbb{Z}/2\mathbb{Z} * \mathbb{Z} = \langle a,b \mid a^2=1\rangle$ | | 2-sheeted covering exists? | Yes | | Covering space | $S^2 \vee S^1 \vee S^1$ |