Suppose $X$ is a manifold of dimension $n$ and $f:Y \to Z$ is an $n-$connected map. Then I want to show that given any map from $g:X \to Y$, the composite map $f \circ g$ is nullhomotopic.
Definition of $n$-connected map: A map is said to be $n$-connected if it induces isomorphisms on all homotopy groups in degree less than $n$ and epimorphisms in degree $n$.
I do not know where to start. If anyone can give some hints as to how $n-$connectedness gets used it would be great. Thanks.
The result is not true. The identity of $X$, $\operatorname{id}_X : X \to X$, is $n$-connected (in fact it's $k$-connected for all $k$). But it is generally not true that for any map $g : X \to X$, the composite $\operatorname{id}_X \circ g = g$ is nullhomotopic: simply consider a non-contractible manifold $X$, then its identity is not nullhomotopic...