Consider by $S_g := T \# T \# ... \# T$ the $g$ times connected sum of tori $T$. Obviously since it is a compact $2$-manifold in light of the famous classification of compact $2$-manifolds $S_g$ is exactly the oriented manifold with genus $g$.

Now the question: I want to know how to prove that there don't exist a $1$-dimensional cellular complex $C_k$ with $rank_{\mathbb{Z}}H_1(C_k, \mathbb{Z}) = k > g$ such that $C_k$ is a retract of $S_g$?

My attempts:

Let assume that it would be true then there exist an inclusion $i: C_k \to S_g$ and a retraction $r: S_g \to C_k$ with $r \circ i = id_{C_k}$. Since $S_g$ path connected then also $C_k =r(S_g)$ so $C_k$ contains as a $1$-dim cellular complex only one $0$-cell.

Therefore $C_g$ can wlog assumed as a bouquet of $k$ $S^1$-circles.

We have therefore following fundamental groups: $\pi_1(S_g) = \langle x_1,y_1, ..., x_g, y_g \rangle / (\prod_{i=1} ^g [x_i, y_i])$ and $\pi_1(C_k) = \langle z_1, ..., z_g \rangle$ is a free group with $g$ free generators. Futhermore, the homology just abelianize the homotopy groups therefore it also don't lead me to desired result.

Up to now I don't find a way to conclude a contradiction.

  • $\begingroup$ When you say retract, do you mean deformation retract ? If yes, the induced map on homology is an isomorphism, which is not possible. $\endgroup$ Nov 4, 2018 at 23:09
  • $\begingroup$ @NicolasHemelsoet: No no, just the "classical" one, so the only condition is that $i \circ r = id_{C_k}$ $\endgroup$
    – user267839
    Nov 4, 2018 at 23:15
  • $\begingroup$ What does $H_1(C_k;\Bbb Z) = k$ mean? Do you mean $\Bbb Z^k$ there? $\endgroup$
    – user98602
    Nov 4, 2018 at 23:33
  • $\begingroup$ @Mike Miller: Not exactly, I meant the rank of $H_1(C_k, \mathbb{Z})$. In generally it could contain torsion. I fixed it. $\endgroup$
    – user267839
    Nov 4, 2018 at 23:46
  • 1
    $\begingroup$ @Neal: Yes, that's true. The classification I mentioned above is of course up to homeomorphisms. But after having study homotopy theory for some time one tends to oversee the finesse to distinguish between "identical" and "up to blabla" :) $\endgroup$
    – user267839
    Nov 5, 2018 at 0:15

1 Answer 1


The map $r_*: H^*(C_k;\Bbb R) \to H^*(S_g;\Bbb R)$ is an injective ring homomorphism, by the fact that it's a retraction.

Poincare duality says that the cup-product pairing on $H^1(S_g;\Bbb R)$ is nondegenerate; another way of phrasing it is that $H^1(S_g;\Bbb R)$ is a symplectic vector space.

A subspace on which the cup-product is trivial is called an isotropic subspace. It is a standard (linear algebra) theorem that an isotropic subspace is of dimension at most half the dimension of the vector space itself. In particular, because the product on $H^1(C_k;\Bbb R)$ is trivial, and $r_*$ is a ring homomorphism, its image is an isotropic subspace. So the image is of rank at most $\dim H^1(S_g;\Bbb R)/2 = g$.

The maps on real cohomology are just the maps in integral cohomology tensored with $\Bbb R$, so you find the same result at the level of integral cohomology rings. In particular, $H^1(C_k;\Bbb Z)$ must have rank at most $g$.


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