# $\pi_2(X)$ defined with closed surfaces instead of spheres

If $X$ is a topological space, the group $\pi_2(X)$ is defined by the set of maps $S^2 \to X$ up to homotopy, where the sum of two maps is the natural map defined on the connected sum of the spheres.

Has been defined / studied the group defined as follows?

Consider the set of maps: $S_g \to X$, where $S_g$ is the closed surface of genus $g$, where $g$ varies in $\mathbb N$.

Given maps $f_1 \,: S_{g_1} \to X$ and $f_2 \,: S_{g_2} \to X$, we define the map $f_1 \# f_2 \,: S_{g_1} \# S_{g_2} \approx S_{g_1 + g_2} \to X$ in the obvious way (the map $f_1 \# f_2$ may be well defined only up to homotopy. Moreover, basepoints should be considered).

The identity of the group is the set of all maps $S_g \to X$ that are homotopically constant. Given $f \, : S_g \to X$, it is possible to define a map $f' \,: S_g \to X$ such that $f \# f'$ is homotopically constant (analogously as in the definition of the inverse in $\pi_2(X)$). Given maps $f_1 \, : S_{g_1} \to X$ and $f_2 \,: S_{g_2} \to X$, we say that $f_1 \sim f_2$ if $f_1 \# f_2$ is homotopically constant (i.e., it represents an element of the unity class of the group).

I think that a similar group could be defined in higher dimensions as well!

EDIT: as pointed out by Miller, the definition above is obviously wrong ($f'$ defined as above does not provide an inverse element). So I ask the related question:

Is it possible to savage the definition above (had corrected version of this "group" being studied?)

• Your claim about the existence of an $f'$ is incorrect. Let $f$ be the identity map of any positive genus surface to itself. Then even the map from the punctured surface to the surface is not null-homotopic; look at homology. Then the map from the punctured surface factors through $f \# f'$.
– user98602
Dec 18, 2016 at 21:50
• The bordism group shares some similarities with your idea Dec 18, 2016 at 21:56

The correct version of the groups you are trying to define has already been defined and studied. The group we are speaking about are called $\textbf{bordism groups}$.

Let us restrict our attention to smooth manifolds, in the topological case you can give similar definitions but it is impossible to prove theorems I think.

Define $\Omega_2(X)$ as follows. Consider the set of pairs $(\Sigma, f)$ with $f: \Sigma \to X$ smooth map from a surface $\Sigma$ to $X$, and impose the equivalence relation $(\Sigma_0, f_0) \sim (\Sigma_1, f_1)$ if there exists a 3-manifold with boundary $Y$ and a smooth map $F:Y \to X$ such that $$\partial Y = \Sigma_0 \cup -\Sigma_1 \ \ \ \text{ and }\ \ \ F|_{\partial Y}= f_0 \cup f_1 \ .$$
We define the sum operation on $\Omega_2(X)$ by setting $$(\Sigma_0, f_0) \# (\Sigma_1, f_1)= (\Sigma_0 \cup \Sigma_1, f_0 \cup f_1).$$

This is at the second bordism group of $X$. I let you guess how to define the higher groups $\Omega_n(X)$ and to prove that these groups satisfy large part of the basic properties of an homology theory. In fact, in some lucky cases you can also prove that $\Omega_n(X) \simeq H_n(X)$. For more about this, look at the last chapter of this book:

http://www.maths.ed.ac.uk/~aar/papers/diecktop.pdf

I would also like to remark that by introducing the possibility to have some singularities in the manifolds involved in this set up, you can define homology groups by using a this construction.

• Notice that two pairs $(\Sigma, f_0)$ and $(\Sigma, f_1)$ with $f_0$ and $f_1$ homotopic are equivalent under the relation $\sim$ Dec 19, 2016 at 1:29