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.
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.