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The fundamental group operation can be considered as a functor $\pi_1:\textbf{Top}\to\textbf{Group}$.

Meanwhile, we know that for any group $G$, there exists a topological space (CW complex) such that $\pi_1(X_G)=G$. My question is: Given a group homomorphism $\varphi:G\to H$, is there always exists a continuous map of CW complexes $f_\varphi:X_G\to X_H$?

I tried to look at Hatcher, but seems nothing like this was mentioned. Thanks in advance in answering.

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    $\begingroup$ Could Hatcher's Proposition 1B.9 be something like what you're looking for? $\endgroup$
    – Elliot Yu
    Aug 18, 2021 at 20:57
  • $\begingroup$ Not quite, Theorem 1B.9 is only for $K(G,1)$ space, meaning fundamental group is fixed to be $G$. I would like to consider arbitrary pair of groups $G$, $H$. $\endgroup$
    – Ivan So
    Aug 18, 2021 at 21:01
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    $\begingroup$ Yes, but couldn't you just choose your spaces $X_G$ and $X_H$ to be $K(G,1)$ and $K(H, 1)$? $\endgroup$
    – Elliot Yu
    Aug 18, 2021 at 21:03
  • $\begingroup$ Oh I see, the theorem just requires the first topological space to be CW complex. However, it is I wanted a theorem where there is no restriction on $Y$. As $Y\in K(G,1)$ implies it has to have a contractible universal covering. $\endgroup$
    – Ivan So
    Aug 18, 2021 at 21:08
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    $\begingroup$ Is the question essentially whether every homorphism $\pi_1(X)\rightarrow\pi_1(Y)$ arises from a map $f:X\rightarrow Y$ for any spaces $X$ and $Y$? Or do you want an existential quantifier on one or both of $X$ and $Y$? $\endgroup$ Aug 19, 2021 at 1:39

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This is not true for general $Y$ with $\pi_1(Y)=H$. The obstruction class is in $H^3(X,\pi_2Y)$.***

To say a bit more, given a homomorphism $\rho:G \to H$, (after contracting $1$-skeletons) define maps $f_1:X^{(1)} \to Y$ by mapping generators for the fundamental group to generators. Since we have a group homomorphism, relations are satisfied, and in $X,Y$ such relations are $2$-cells, and so there is no problem extending this to a cellular map $f_2:X^{(2)} \to Y$.

After this, however, we are out of such luck. If $X$ is the presentation complex for $G$, then this is fine, but otherwise we have an obstruction cocycle that lives in $H^3(X,\pi_2 Y)$.

As mentioned before, when $Y=K(G,1)$, we know that $\pi_2(Y)=0$ and terefore such an obstruction doesn't exist, so we can extend to $X^{(3)}$, and to extend further we have another class in $H^4(X,\pi_3(Y))$, and so on, but $\pi_i(Y)=0$ for all $i>1$, and thus all of these obstructions vanish. As mentioned in the comments, this can be proven without this language (although basically the same idea) in Hatcher 1B.9.

A concrete counter-example would be finding a continuous map $f:\mathbb RP^3 \to \mathbb RP^2$ inducing the identity on $\mathbb Z/2$. One shows that this is impossible (without obstruction theory) by using the cup product on cohomology (see the proof of the Borsuk Ulam theorem here.

This news is not so bad, since the assignment $G \to K(G,1)$ can be made functorial, and in this case, we can always realize group homomorphisms as continuous maps between some choice of spaces.


***We should assume here that $Y$ is a simple space to make this more faimiliar, but we can in fact have an equivalent formulation by using local coefficients for cohomology.

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  • $\begingroup$ I see what you meant. So it's like if the concerned pair of topological space are both 2-complex, then it is possible. But if it is of dimension >=3, then not always. Am I right? $\endgroup$
    – Ivan So
    Aug 21, 2021 at 20:05
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    $\begingroup$ yes, its actually enough for $X$ to be a 2-complex, or alternatively for the higher homotopy of $Y$ to vanish. The real condition is the obstruction cocycle vanishing altogether in all dimensions. $\endgroup$ Aug 22, 2021 at 2:32

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