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If $f : X \to Y$ is a continuous function and we fix base points $x_0 \in X$ and $y_0 = f(x_0)\in Y$, then $f$ induces a group homomorphism $$f_*: \pi_1(X, x_0) \to \pi_1(Y, y_0)$$ between the fundamental groups of $(X, x_0)$ and $(Y,y_0)$. I was just wondering if there is some kind of reciprocal property, that is:

If $\psi : \pi_1(X,x_0) \to \pi_1(Y,y_0)$ is a homomorphism, is it true that $\psi = f_*$ for some continuous function $f: X \to Y$?

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  • $\begingroup$ I'm having trouble finding an obvious function for this one, although I don't know whether it's an actual counterexample: Let $X$ be the torus and $Y$ be the circle. Let $\psi$ be the map $\pi_1(X, x_0)\to \pi_1(Y, y_0)$ that sends the $(a, b)$ torus knot to $a-b$. $\endgroup$
    – Arthur
    Oct 25, 2017 at 9:17
  • $\begingroup$ @Arthur That would be a nice counterexample if it is actually true! Thank you :) $\endgroup$
    – user313212
    Oct 25, 2017 at 9:37
  • $\begingroup$ @Arthur, let $m:S^1\times S^1\rightarrow S^1$ be the multiplication and $\iota:S^1\rightarrow S^1$ the inverse. Then $f=m\circ(1\times\iota)$ is the map you need. $\endgroup$
    – Tyrone
    Oct 25, 2017 at 9:40

2 Answers 2

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Not in general, there are obstructions.

A discussion about this interesting problem is contained in MathOverflow question 166153.

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  • $\begingroup$ Thank you for the reference, I didn't see that MathOverflow question D: $\endgroup$
    – user313212
    Oct 25, 2017 at 9:29
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The question in the title of your post and the question in the body are different. If you start with spaces $X$ and $Y$ and a homomorphism from $\pi_1(X,x_0)$ to $\pi_1(Y, y_0)$, then there are obstructions, as Francesco says. And this is probably what you had in mind.

However, it might be worth pointing out that if you start with an abstract homomorphism $\phi:G\to H$, then you can construct spaces $X$ and $Y$ with $\pi_1(X,x_0)=G$ and $\pi_1(Y,y_0)=H$, and a continuous function $f:X\to Y$ such that $f_*=\phi$. Namely, you can take X and Y to be Eilenberg Maclane spaces for G and H, respectively. See Hatcher page 90 for details.

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  • $\begingroup$ Actually, it suffices to choose $Y$ to be an Eilenberg Maclane space for $H$. $X$ just needs to be a connected CW complex with fundamental group $G$. Also, the map $f$ is unique up to homotopy! $\endgroup$
    – beanstalk
    Oct 26, 2017 at 1:49

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