Suppose $Y$ is an $(n-1)$-connected space, $n>2$, so we have Hurewicz isomorphisms $\pi\_n(Y)\cong H\_n(Y)$ and $\pi\_{n-1}(\Omega Y)\cong H\_{n-1}(\Omega Y)$. Let a map $\alpha\colon X\to\Omega Y$ be given. Naturally it induces a map $\beta\colon X\times S^1\to Y$. I want to show the following diagram is commutative:
$$\require{AMScd}
\begin{CD}
H\_{n-1}(X) @>\times[S^1]>> H\_n(X\times S^1)\\
@V\alpha\_\*VV @V\beta\_\*VV \\
H\_{n-1}(\Omega Y) @<\cong<< H\_n(Y).
\end{CD}
$$


Since $\Omega Y$ is path connected, we may homotope $\alpha$ to a based map. Then $\beta$ factors though the reduced suspension $\Sigma X$. If $X=S^{n-1}$ is a sphere, the commutativity would then follow from tracking down the definition of $\pi\_n(Y)\xrightarrow{\cong}\pi\_{n-1}(\Omega Y)$. However I don't know how this helps for the general case.


One can also phrase the question in cohomology in the obvious way. (In particular the cross product $\times[S^1]$ will be replaced by the slant product $/[S^1]$.)



Induced map in homology for a map to a loop space
The general case does follow from the case $Y=S^n$.


Without loss of generality $X=\Omega Y$ and $\alpha$ is the identity. Now the statement comes down to the assertion that the composition
$$
\pi\_{n-1}\Omega Y\to H\_{n-1}\Omega Y\to H\_n(\Omega Y\times S^1)\to H\_n Y
$$
corresponds to the Hurewicz map $\pi\_nY\to H\_nY$ via a natural isomorphism between $\pi\_{n-1}\Omega Y$ and $\pi\_nY$. This is true for all $Y$, and by naturality it is enough to prove it for $Y=S^n$.


