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Suppose $X$ is a CW-complex such that there is a stable splitting of $X$ into wedge sum $$ \Sigma^t X\cong \bigvee _{k=1}^\infty Y_k. $$

(1). Does this imply $$ X\to \Sigma^tX\to \bigvee _{k=1}^\infty Y_k\to Y_k $$ induce an epimorphism on homology $$ H_*(X)\to H_*(Y_k)? $$

(2). Can we construct a map $$ Y_k\to \bigvee _{k=1}^\infty Y_k\to \Sigma^tX \to X $$ such that the map induces a monomorphism on homology?

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    $\begingroup$ What's the map from $X$ to its suspension? $\endgroup$
    – Mark Grant
    Commented May 25, 2015 at 9:48
  • $\begingroup$ I want the map to be inclusion $X\to X\wedge S^t$. Is it valid? $\endgroup$
    – Shiquan Ren
    Commented May 25, 2015 at 11:06
  • $\begingroup$ @RSQ That map factors through the inclusion of $X$ into the cone on $X$ and is therefore nullhomotopic. Consider the equatorial inclusion of one $S^n \to S^{n+1}$. $\endgroup$
    – Sean Tilson
    Commented May 27, 2015 at 9:01

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Rather than thinking about maps $X\to \Sigma^tX$ or $\Sigma^tX\to X$ you should just pre- or post-compose with the suspension isomorphism.

For instance, $$ H_*(X) \cong \tilde{H}_{*+t}(\Sigma^t X) \cong \tilde{H}_{*+t}\left(\bigvee_{k=1}^\infty Y_k\right) \twoheadrightarrow \tilde{H}_{*+t}(Y_k) $$ is an epimorphism (the last map is an epimorphism because it's induced by a retraction). So we get an epimorphism $H_*(X)\to H_*(Y_k)$ up to a degree shift.

Similarly you'll get a monomorphism $H_*(Y_k)\to H_*(X)$ up to a degree shift.

Note that the inclusion $X\to \Sigma^t X$ is always null-homotopic.

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