Calculating the (de-Rham) cohomology of a tee connector (Picture), I got $H^0=R,H^1=R^2,H^2=0$.
Furthermore, just from looking at it, I assume the tee connector is homotopic to a circle with an arc connecting two antipodal points.
Is it always true that an $n$-dimensional manifold with trivial cohomology groups for $k_0< k \leq n$ is homotopic to a $k_0$-dimensional manifold?
No; for instance, there are acyclic spaces (i.e. with vanishing cohomology in dimension $>0$) which are not contractible (i.e. not homotopic to a $0$-dimensional manifold). For instance, the Poincaré sphere with a point deleted from it is acyclic, but it is not contractible because it has nonvanishing $\pi_1$. (The $\pi_1$ of the non-punctured Poincaré sphere is the binary isocahedral group of order $120$, whose abelianization is trivial. Therefore, $H^1$ vanishes, and the cohomology is concentrated in degree $0$ and in the top degree $3$. Deleting a point then kills the top cohomology.)