Homotopy
Non-surjective maps $X\to X^n$ are nullhomotopic (Tag P-KIIUK)
- Show that if \(X\mapsvia{f} X^n\) is not surjective, then \(f\) is nullhomotopic.
Solution
<1>1. The notation \(X^n\) is not defined in the statement, so the target space is not determined.
<1>2. For spheres, if \(f:S^n\to S^n\) is not surjective, then \(f\) is null-homotopic.
Choose \(p\in S^n\setminus f(S^n)\). Then \(f\) factors through \(S^n\setminus\{p\}\cong\mathbb R^n\), which is contractible. Hence the factorization, and therefore \(f\), is null-homotopic.
\(\pi_1\)
$\pi_1(S^1\vee S^1)$ and $\pi_1(S^1\times S^1)$ (Tag P-FHLRF)
Compute \(\pi_1(S^1 \vee S^1)\)
Compute \(\pi_1(S^1 \cross S^1)\)
Solution
<1>1. The figure-eight satisfies \[\boxed{\pi_1(S^1\vee S^1)\cong F_2.}\]
Apply van Kampen to neighborhoods of the two circle summands whose intersection is contractible. The result is the free product \(\mathbb Z*\mathbb Z\).
<1>2. The torus satisfies \[\boxed{\pi_1(S^1\times S^1)\cong\mathbb Z^2.}\]
Fundamental groups commute with products, so \(\pi_1(S^1\times S^1)\cong\pi_1(S^1)\times\pi_1(S^1)\cong\mathbb Z^2\).
$\pi_1(S^1\vee S^1)$ (Tag P-24CMJ)
Compute the fundamental group \(\pi_1(S^1 \vee S^1, x_0)\) of the wedge sum of two circles (the figure eight space).
Solution
Let \(X=S^1_a\vee S^1_b\) with wedge point \(x_0\).
<1>1. Choose points \(p_a\in S^1_a\setminus\{x_0\}\) and \(p_b\in S^1_b\setminus\{x_0\}\), and put \[ U=X\setminus\{p_b\}, \qquad V=X\setminus\{p_a\}. \] Then \(U\simeq S^1_a\), \(V\simeq S^1_b\), and \(U\cap V\) is contractible.
<1>2. Therefore \[ \pi_1(U,x_0)\cong\mathbb Z, \qquad \pi_1(V,x_0)\cong\mathbb Z, \qquad \pi_1(U\cap V,x_0)=1. \]
<1>3. Seifert–van Kampen gives \[ \pi_1(S^1\vee S^1,x_0) \cong \mathbb Z*\mathbb Z \cong F_2. \] Thus the figure eight has free fundamental group on the two circle loops.
$\pi_1(S^1 \times S^1)$ (Tag P-CEVX4)
- Compute \(\pi_1(S^1 \cross S^1)\)
Solution
<1>1. For path-connected spaces, fundamental groups commute with products: \[\pi_1(X\times Y,(x_0,y_0))\cong\pi_1(X,x_0)\times\pi_1(Y,y_0).\]
A loop in a product is exactly a pair of loops, and homotopies are likewise coordinatewise.
<1>2. Since \(\pi_1(S^1)\cong\mathbb Z\), \[\boxed{\pi_1(S^1\times S^1)\cong\mathbb Z\times\mathbb Z.}\]
Apply <1>1 to the two circle factors.
Surfaces
Even-degree orientable covers of non-orientable manifolds, and $\pi_1$ without index-$2$ subgroups (Tag P-VTB25)
Let \(M\) be a connected topological manifold. (1) Prove that if \(\widetilde{M}\) is an orientable connected manifold and \(p: \widetilde{M} \to M\) is a \(k\)-fold covering map onto a non-orientable manifold \(M\), then the covering degree \(k\) must be even (or infinite). (2) Prove that if \(\pi_1(M)\) has no subgroup of index \(2\), then \(M\) must be orientable.
Solution
Let \[ w:\pi_1(M)\longrightarrow\{\pm1\}\cong\mathbb Z/2 \] be the orientation character. Since \(M\) is connected, it is orientable exactly when \(w\) is trivial.
<1>1. Suppose \(M\) is non-orientable and \(p:\widetilde M\to M\) is a connected covering with \(\widetilde M\) orientable. Put \[ H=p_*\pi_1(\widetilde M)\le\pi_1(M). \] Then \[ H\subseteq\ker w. \]
A loop in \(\widetilde M\) preserves the orientation of \(\widetilde M\). Its projection therefore has trivial orientation monodromy in \(M\), so its class lies in \(\ker w\).
<1>2. Since \(M\) is non-orientable, \(w\) is surjective, so \[ [\pi_1(M):\ker w]=2. \] Thus, when the covering has finite degree, \[ k=[\pi_1(M):H] =[\pi_1(M):\ker w]\,[\ker w:H] =2[\ker w:H], \] so \(k\) is even. If \([\ker w:H]\) is infinite, then the covering has infinitely many sheets.
<1>3. If \(\pi_1(M)\) has no subgroup of index \(2\), then \(M\) is orientable.
If \(M\) were non-orientable, the surjective orientation character would have kernel of index \(2\), contradicting the hypothesis.
Orientable covers of non-orientable manifolds have even or infinite degree (Tag P-D6JY5)
- Show that if \(M^\text{orientable} \mapsvia{\pi_k} M^\text{non-orientable}\) is a \(k\dash\)fold cover, then \(k\) is even or \(\infty\).
Solution
<1>1. Let \(p:\widetilde M\to M\) be a connected orientable covering of a connected nonorientable manifold, and let \(H=p_*(\pi_1\widetilde M)\le\pi_1(M)\).
The number of sheets is the index \([\pi_1(M):H]\).
<1>2. The orientation character \(w_1:\pi_1(M)\to\mathbb Z/2\) is nontrivial and vanishes on \(H\).
Nonorientability makes \(w_1\) onto. A loop in \(\widetilde M\) preserves the chosen orientation there, so its projected loop has trivial orientation character.
<1>3. Therefore \(H\subseteq\ker w_1\), and \[[\pi_1(M):H]=2[\ker w_1:H]\] whenever the latter index is finite.
The kernel of the surjective orientation character has index \(2\); use multiplicativity of subgroup indices.
<1>4. Thus an orientable cover of a nonorientable manifold has even finite degree or infinite degree.
This follows directly from <1>3.
A manifold is orientable if $\pi_1$ has no subgroup of index $2$ (Tag P-6HH2P)
Let \(M\) be a connected topological/smooth manifold. Prove that if the fundamental group \(\pi_1(M)\) has no subgroup of index 2, then \(M\) is orientable.
Solution
<1>1. A connected manifold \(M\) has an orientation character \[ w:\pi_1(M,x_0)\longrightarrow\{\pm1\}\cong\mathbb Z/2, \] which records whether transport of a local orientation around a loop preserves or reverses it.
<1>2. The manifold is orientable if and only if \(w\) is trivial. <2>1. If \(M\) is orientable, a global orientation is preserved along every loop, so \(w=1\). <2>2. Conversely, if \(w\) is trivial, a choice of local orientation at \(x_0\) transports independently of path to every point, producing a global orientation.
<1>3. Suppose \(M\) were nonorientable. Then \(w\) would be nontrivial. Since its codomain has order \(2\), nontriviality makes \(w\) surjective, and therefore \[ \ker w\le\pi_1(M) \] has index \(2\).
<1>4. This contradicts the hypothesis that \(\pi_1(M)\) has no subgroup of index \(2\). Hence \(M\) is orientable.
<1>5. Equivalently, \(\ker w\) is the subgroup corresponding to the connected orientation double cover of a nonorientable manifold.