Two topological spaces $X$ and $Y$ are homotopic if there exists continuous $f: X \to Y$ and $g: Y\to X$ such that $f\circ g$ is homotopic to $Id_Y$ and $g\circ f$ homotopic to $Id_X$ (regardless any embedding of $X$ and $Y$ in a bigger topological space). This leads to the definition of a simply connected space; a space where any two loops are homotopic and I am confused:

1- Any two circles are homotopic (because they are diffeomorphic), So what means non homotopic circles on the torus $T^2$? hence my problem: is a homotopy a relative concept?

2- If I take as a definition of the simple connectedness of a manifold $M$, the vanishing of its de Rham cohomology space $H^1(M)=0$. One can easily show that the sphere $S^2$ is simply connected. By Stereographic projections by $S^2=U\cup V$ with $U$ and $V$ homeomorphic to $\mathbb{R}^2$ (thus contractible) and $U\cap V$ is homotopic to the circle $S^1$. Using the long exact sequence of Mayer-Vietoris: $$0\to H^0(S^2)\to H^0(U)\oplus H^0(V)\to H^0(U\cap V)\to H^1(S^2)\to 0$$ so $1-2+1-\dim H^1(S^2)=0$.

How to show that two loops in $S^2$ are homotopic using the fact that $H^1(S^2)=0$? I know that if a $1$-differential form is exact then its integral over all loop is zero. Is the converse true?

Thanks for any help.

• Homotopy is usually applied to maps, not to spaces. In homotopic topology, a “circle on X” is understood as a map from S¹ to X. – Incnis Mrsi Dec 2 '14 at 18:28
• The property you wrote down (there exist continuous maps $f: X \to Y$ and $g: Y\to X$ such that $f\circ g$ is homotopic to $\operatorname{Id}_Y$ and $g\circ f$ homotopic to $\operatorname{Id}_X$) is what it means to say that $X$ and $Y$ are homotopy equivalent, not homotopic. – Jack Lee Dec 2 '14 at 21:59
• Are you sure you are supposed to show that $S^2$ is simply-connected just using $H^1_{\operatorname{dR}}(S^2)=0$? The only arguments I'm familiar with use facts (e.g. that simple-connectedness is equivalent to $H^1=0$ for finite polyhedra in $S^3$) that are far more powerful than, say, van Kampen's theorem or basic facts about CW complexes (which can easily show that $S^2$ is simply-connected). – Kyle Dec 2 '14 at 22:35

I'm going to echo Incnis Mrsi's comment: The first concept you describe is called homotopy equivalence of spaces. Two spaces are then homotopy equivalent if there are compositions of continuous maps $X \overset{f}{\to} Y \overset{g}{\to} X$ homotopic to $\operatorname{id}_X$ and $Y \overset{g}{\to} X \overset{f}{\to} Y$ homotopic to $\operatorname{id}_Y$. As for your specific questions:

1. When we say "a circle in the torus $T^2$", we usually mean a continuous map $f: S^1 \to T^2$. Two homotopic circles in the torus are then two continuous maps $f,g:S^1 \to T^2$ such that there is another continuous map $F : S^1 \times [0,1] \to T^2$ with $F_0=f$ and $F_1=g$. The subspaces $f(S^1)\subset T^2$ and $g(S^1)\subset T^2$ may not be homotopy equivalent even if $f$ and $g$ are homotopic, and vice versa. For example, all loops $f:S^1 \to \mathbb{R}^2$ are homotopic to the constant map $f: S^1 \to \{0\} \subset \mathbb{R}^2$ because $\mathbb{R}^2$ is contractible. But the unit circle $S^1 \subset \mathbb{R}^2$ is certainly not homotopy equivalent to the origin $\{0\} \subset \mathbb{R}^2$.

2. While it is true that a manifold $M$ that is simply-connected in the standard sense, i.e. in which all loops $S^1 \to M$ are homotopic, will have $H^1 _{\operatorname{dR}}(M)=0$, the converse is false. There are non-simply-connected manifolds whose first de Rham cohomology groups are zero. For example, if you know about (singular) (co)homology with coefficients (which I'll denote by $H_*$ and $H^*$), you can use $$H^1_{\operatorname{dR}}(M) \cong H^1(M;\mathbb{R}) \cong H^1(M;\mathbb{Z}) \otimes_{\mathbb{Z}} \mathbb{R} \cong \operatorname{Hom}_\mathbb{Z}(H_1(M;\mathbb{Z}),\mathbb{R})$$ to compute $H^1_{\operatorname{dR}}(\mathbb{RP}^3)\cong (\mathbb{Z}/2\mathbb{Z})\otimes \mathbb{R}=0$. But $\pi_1(\mathbb{RP}^3)\cong \mathbb{Z}/2\mathbb{Z} \neq 0$, so $\mathbb{RP}^3$ is not simply-connected.

Update: In your comment below, you asked if the map $\varphi: H^1(M) \to \operatorname{Hom}(\pi_1(M),(\mathbb{R},+))$ defined by $(\varphi(\omega))(\gamma)=\int_\gamma \omega$ is an isomorphism. The answer is yes (at least for compact manifolds), but you may not yet have learned the tools necessary to see this. To see that it is injective, we'll show that the kernel is trivial: Suppose $(\varphi(\omega))(\gamma)=\int_\gamma \omega=0$ for all $\gamma: S^1 \to M$. This implies that $\omega$ is an exact form $df$, where $f$ can be constructed by setting $f(x_0)=0$ for some chosen $x_0 \in M$ and then letting $f(x)=\int_c \omega$ where $c$ is any curve from $x_0$ to $x$. Thus $[\omega]=0$ in $H^1(M)$, and $\varphi$ is injective. A proof of surjectivity would probably require tools that you have not yet learned, such as the fact that singular homology $H_1(M;\mathbb{Z})$ is isomorphic to the abelianization of $\pi_1(M)$, i.e. the quotient $\pi_1(M)/[\pi_1(M),\pi_1(M)]$ where $[\pi_1(M),\pi_1(M)]$ is the "commutator subgroup" generated by elements like $aba^{-1}b^{-1}$.

Unfortunately, this isomorphism might not help prove $\pi_1(S^2)=0$. This is because a nontrivial group $G\neq 0$ can have $\operatorname{Hom}(G,(\mathbb{R},+))=0$, such as $G=\mathbb{Z}/2\mathbb{Z}$. If you can show that $\pi_1(S^2)$ is abelian and torsion-free, then $\operatorname{Hom}(\pi_1(S^2),(\mathbb{R},+))\cong H^1(S^2)=0$ will imply $\pi_1(S^2)=0$, but this is way harder than showing $\pi_1(S^2)=0$ directly.

• @ squirrel: Many thanks for your nice answer, the terminology is now clear. I am not familiar with singular (co)homology with coefficients. I use only the fact that forms on the projective space are isomorphic to $\mathbb{Z}_2$-invariant forms on the sphere to show that $H^k(\mathbb{RP}^n)\simeq\left(H^k(S^n)\right)^{\mathbb{Z}_2}=0$ for $n\geq2$. How to prove $\pi_1(M)=0\longrightarrow H_{dR}^1(M)=0$ ? the map $\varphi : \omega\mapsto\left(\gamma\to\int_\gamma \omega\right)$, from $H^1(M)$ to $\{homomorphisms : \pi_1(M)\to(\mathbb{R},+)\}$ is it an isomorphism? – user56980 Dec 4 '14 at 11:13
• @user56980: Just updated the post in response to your comment. – Kyle Dec 9 '14 at 16:29
• Many thans squirrel! Yes, if $G$ is a finite group than $\operatorname{Hom}(G,(\mathbb{R},+))=0$. I'm bad in group theory (free groups, amalgamated product ...) but I can use a result (from Van Kampen) which states that the union of two simply connected, whose intersection is path connected is simply connected. May I ask you some references for group theory necessary for singular homology and references for cohomology with coefficients. Thanks! – user56980 Dec 10 '14 at 21:18
• @user56980: Indeed, the argument you mentioned, namely applying Van Kampen's theorem to the decomposition $S^2 = (S^2 \setminus \{\text{north pole}\}) \cup (S^2 \setminus \{\text{south pole}\}$, is the easiest way to prove $\pi_1(S^2)=0$. I think most people would recommend Algebraic Topology by Allen Hatcher (freely available online). Hatcher covers free groups and amalgamated products in Chapter 1, where he discusses the fundamental group and covering space theory. – Kyle Dec 11 '14 at 7:36
• That's great, thanks a lot squirrel! – user56980 Dec 11 '14 at 9:48