Relative homology of free loop space with respect to constant loops Let $Q$ be a closed manifold with $\dim Q\geq2$ and let $\Lambda_0Q$ be the connected component of the free loop space of $Q$ whose elements are contractible loops. I am looking for conditions on the homotopy groups $\pi_k(Q)$ ensuring that the relative homology $H(\Lambda_0Q,Q)$ with respect to the space of constant loops $Q$ is non-zero.
Using the relative Hurewicz theorem, I can see that $H(\Lambda_0Q,Q)\neq0$ if $Q$ is $2$-connected and I was interested in a sharper result. Is it true that it is enough to assume that $Q$ is $1$-connected? What about a non-simply connected manifold $Q$ having $\pi_k(Q)\neq0$ for some $k\geq2$?
 A: Let me tackle the case that $Q$ is $1$-connected, but not $2$-connected. Because $Q$ is $1$-connected we have that $\Lambda_0Q=\Lambda Q$, as all loops are contractible.
Two sequences are relevant here: The long exact sequence for homotopy groups of the free loop space fibration $\Omega Q\rightarrow \Lambda Q\xrightarrow{\mathrm{ev}} Q$, and the long exact sequence in homology of the pair $(\Lambda Q,Q)$.
The evaluation of the basepoint $\mathrm{ev}$ has a section: It sends a point in $q\in Q$ to the constant loop at $q$. This means that the long exact sequence in homotopy groups splits into split short exact sequences:
$$
0\rightarrow \pi_k(\Omega Q)\rightarrow \pi_k(\Lambda Q)\rightarrow \pi_k(Q)\rightarrow 0
$$
Now $\pi_k(\Omega Q)=\pi_{k+1}(Q)$, and all groups in sight are abelian because $Q$ is connected. The splitting lemma then shows that $\pi_k(\Lambda Q)\cong\pi_{k+1}(Q)\oplus \pi_k(Q)$. Similarly, the long exact sequence in homology breaks down and
\begin{equation}
H_k(\Lambda Q)\cong  H_k(Q)\oplus H_k(\Lambda Q,Q)
\end{equation}
By assumption $\pi_2(Q)\not=0$. By Hurewicz (note that $\pi_1(\Lambda Q)$ is abelian) we have $H_1(\Lambda Q)\cong \pi_1(\Lambda Q)\cong \pi_2(Q)\not=0$. As $H_1(Q)\cong\pi_1(Q)=0$, the last displayed equation gives that $H_1(\Lambda Q,Q)\not=0$.
