# Classifying continuous maps from closed 2-manifolds to various closed manifolds

I believe my question should be simple. The question is more physically oriented and originated from one of Witten's papers, "On Holomorphic Factorization of WZW and Coset Models", where he considered a continuous map, $$\phi: \Sigma\rightarrow M\,,$$ where $\Sigma$ is a two-dimensional closed surface and $M$ is an arbitrary closed manifold. Witten claimed that, if $$\pi_1(M)=\pi_2(M)=0\,,$$ then the map $\phi$ will be automatically nullhomotopic. Denote the set $\,\mathcal{C}=\{\phi:\Sigma\rightarrow M\}$ as all continuous maps from $\Sigma$ to $M$.

Now my question is simply that if either $\pi_1(M)$ or $\pi_2(M)$ are non-trivial. How can one argue that there would be maps in $\mathcal{C}$ NOT homotopic to identity. More generally, if assume $\Sigma$ is a 2-sphere, can one figure out the homotopy of the configuration space $\,\mathcal{C}$, say $\pi_0(\mathcal{C})$, $\pi_1(\mathcal{C})$ and $\pi_2(\mathcal{C})$ etc.

• "[...] maps in $\mathcal{C}$ NOT homotopic to identity" did you mean not homotopic to the zero/constant map? Identity doesn't make much sense there. Jul 7 '16 at 10:15
• if $\Sigma=S^2$, then $\mathcal{C}$ is just the double-loopspace $\Omega^2 M$, so $\pi_n(\mathcal{C})=\pi_{n-2}(M)$ Jul 7 '16 at 14:46
• sorry, i meant $\pi_n(\mathcal{C})=\pi_{n+2}(M)$ Jul 7 '16 at 19:02
• @BalarkaSen Yes, I should have corrected it as "homotopic to constant map". I have an additional question posted in your answer. Would you please have a check? Thanks! Jul 11 '16 at 11:21

If we know that either $\pi_1(M)$ or $\pi_2(M)$ is nontrivial, indeed there does exist a homotopically nontrivial map $\Sigma \to M$ for any surface $\Sigma \neq S^2, \Bbb{RP}^2$.
If $\pi_1(M) \neq 0$, then choose a representative $f : S^1 \to M$ of a nontrivial homotopy class $[f]$. Let $\Sigma$ be some surface, and let $\Sigma \to T^2$ be the map given by pinching complement of a punctured torus in the connected sum decomposition of $\Sigma$. Then compose with the projection map $T^2 \to S^1$ to get a map $\Sigma \to S^1$. This is clearly nontrivial on $\pi_1$. Consider then the composition $\Sigma \to S^1 \to M$, which is our desired non-nullhomotopic map, as it is nonzero on $\pi_1$.
(Note that the technique doesn't work if $\Sigma$ doesn't have a torus in the connected sum decomposition, i.e., if $\Sigma \cong S^2$. In that case, indeed, $\pi_1 \neq 0$ is not sufficient for there to exit a non-null map from $S^2$: take, e.g., $M = S^1$. For nonorientable $\Sigma$, same argument applies by finding a Klein bottle component in $\Sigma$ instead of a torus, again with the exception of $\Bbb{RP}^2$ - if there was a homotopically nontrivial map $\Bbb{RP}^2 \to S^1$, there would have been a nontrivial homomorphism $\Bbb Z/2 \to \Bbb Z$ at the level of fundamental groups, which is nonsense)
If $\pi_1(M) = 0$ but $\pi_2(M) \neq 0$, consider a representative of some nonzero homotopy class $[f]$ in $\pi_2(M)$, which gives you a non-nullhomotopic map $f : S^2 \to M$. Moreover, since $\pi_1(M) = 0$, under the Hurewicz isomorphism $\pi_2(M) \to H_2(M)$, $[f]$ is sent to a nontrivial homology class $[f^*(S^2)]$ in $M$, hence $f$ is nontrivial in homology. If $\Sigma$ is some higher genus surface, let $\Sigma \to S^2$ be the degree $1$ map given by pinching a complement of a small ball inside a chart in $\Sigma$. The composition $\Sigma \to S^2 \to M$ is non-nullhomotopic, because it's nonzero in $H_2$ (as so are the two maps in the composition). If $\Sigma$ is nonorientable, exact same argument works, except you have to work in homology mod 2.
• Just to add to the picture: If $\Sigma=S^2$ then each map $\Sigma\to M$ is null-homotopic iff $\pi_2(M)=0$. If $\Sigma=RP^2$ and $\pi_1(M)$ has nontrivial 2-torsion and $\pi_2(M)=0$ then there is a homotopically nontrivial map $\Sigma\to M$. The situation when $\pi_2(M)\ne 0$ is more complicated. Jul 8 '16 at 16:26
• @BalarkaSen Thanks for your answer and it's very instructive! In addition, if there exists non nullhomotopic maps in $\mathcal{C}$, does it represent a generator in certain homotopy groups of $\mathcal{C}$ Jul 11 '16 at 11:21
• @Loliphilia Sure, it represents a nontrivial element in $\pi_0(\mathcal{C})$. Elements of $\pi_0(\mathcal{C})$ are in one-to-one correspondence with homotopy classes of elements in $\mathcal{C}$. Jul 11 '16 at 15:01