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I am interested in the following question which I asked at MSE and did not get any clues. How to find three two-dimensional CW complexes $K_1, K_2, K_3$ with non-trivial $\pi_2$ and injective embeddings $f:K_1\rightarrow K_2, g:K_2\rightarrow K_3$, such that the induced homomorphisms $f_*, g_*$ on the level of $\pi_2$ would be trivial?

The question with one embedding is relatively simple. Consider $K_1 = \mathbb{R}P^2, K_2 = \mathbb{R}P^2\cup C$, where $C$ is the 2-dimensional cell which kills $\pi_1$. Then embedding $f: \mathbb{R}P^2 \rightarrow \mathbb{R}P^2\cup C$ has the property that $f_* = 0$. This follows from Hurewicz theorem and the fact that $H_2(\mathbb{R}P^2)=0$.

The problem and the example are taken from R. Mikhailov lectures. This problem is tied with Whitehead asphericity conjecture.

Also, any references which could help to feel this kind of algebraic topology deeper are welcome.

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It would suffice to introduce new generator of $\pi _2$ in $K_2$. So, for example, we can take $$K_1=RP^2,K2=RP^2\cup C \cup RP^2, K_3= RP^2\cup C\cup RP^2\cup C$$ where C is as in your notation.

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    $\begingroup$ Are you sure $K_2\to K_3$ induces a zero map on $\pi_2$? Why? $\endgroup$
    – Grigory M
    Commented Jan 14, 2015 at 8:56
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    $\begingroup$ Indeed, this cannot work: there is no embeding from $RP^2\cup C$ into any 2-dimensional CW complex $Y$ such that the induced map $\pi_2(RP^2\cup C)\to\pi_2(Y)$ is zero. Easier example: there is no embeding from $S^2$ into any 2-dimensional CW complex $Y$ such that the induced map $\pi_2(S^2)\to\pi_2(Y)$ is zero. The reason is that any such embedding $S^2\to Y$ admits a retract $Y\to S^2$ up to homotopy. $\endgroup$
    – André Henriques
    Commented Jan 14, 2015 at 10:17
  • $\begingroup$ I have got almost the same answer on MSE (it was removed), of course, as Andre noticed it cannot work. $\endgroup$
    – Samarkand
    Commented Jan 14, 2015 at 15:21
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    $\begingroup$ More generally, both $K_1$ and $K_2$ must have the property that the Hurewicz map $\pi_2\to H_2$ vanishes, because any inclusion of 2-dimensional complexes is injective on $H_2$. $\endgroup$
    – Eric Wofsey
    Commented Jan 14, 2015 at 18:37

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