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If two space X and Y have isomorphic fundamental group then when can we say that this isomorphism is induced by some map f from X to Y ? and also what can we say about this question when we take higher homotopy group in place of fundamental group ?

Shivani Sengupta
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    If $X$ is an arbitrary connected CW complex and $Y$ is aspherical (contractible universal cover), then you can say that any homomorphism $\pi_1(X) \to \pi_1(Y)$ is induced from a map $X \to Y$. In particular if $X, Y$ are both aspherical, you can say any isomorphism on the fundamental group is induced from a homotopy equivalence. – Balarka Sen Jul 08 '16 at 12:16
  • @BalarkaSen can you please give me some reference how to prove this ... – Shivani Sengupta Jul 08 '16 at 12:31
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    Look at Hatcher's algebraic topology book, section 1.A. – Balarka Sen Jul 08 '16 at 12:32

2 Answers2

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Not in general.

For example, let $X = \mathbb{R}P^n$ and $Y = \mathbb{R}P^m$ for $n > m$. I claim that for any continuous map $f:X\rightarrow Y$, that $f$ induces the $0$ map on $\pi_1 = \mathbb{Z}/2\mathbb{Z}$.

To see this, assume $f$ is non-trivial on $\pi_1$. Since $H_1$ is the abelianization of $\pi_1$, it follows that $f$ induces an isomorphism $H_1(X;\mathbb{Z}/2\mathbb{Z})\rightarrow H_1(Y;\mathbb{Z}/2\mathbb{Z})$. Universal coefficients then implies that $f$ induces an isomorphism $H^1(Y;\mathbb{Z}/2\mathbb{Z})\rightarrow H^1(X;\mathbb{Z}/2\mathbb{Z})$.

Now, $H^\ast(X;\mathbb{Z}/2\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}[x]/x^{n+1}$ and $H^\ast(Y;\mathbb{Z}/2\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}[y]/y^{m+1}$ where both $x$ and $y$ are degree one elements. By the previous paragraph, we have $f^\ast(y) = x$.

But then $$0 = f^\ast(y^{m+1}) = f^\ast(y)^{m+1} = x^{m+1} \neq 0 $$ because $m+1 < n+1$. This contradiction implies $f$ must have induced the $0$ map on $\pi_1$.

One can use a similar argument with $\mathbb{C}P^n$ (respectively $\mathbb{H}P^n$) replacing $\mathbb{R}P^n$ in order to get an example using $\pi_2$ (respectively $\pi_4$).

Jason DeVito - on hiatus
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This is not really an answer, but it's too long for a comment.

I want to observe that while Jason has shown it is not in general true that every homomorphism of fundamental groups comes from a map of spaces, it is true for a particular class of spaces, namely Eilenberg-Mac Lane spaces. That is, $$[K(G,1), K(H,1)] \cong \operatorname{Hom}(G,H)].$$ So, if the two spaces in your question are Eilenberg-Mac Lane, then there is no problem to realizing the homomorphism of fundamental groups.

Now suppose $X$ and $Y$ are arbitrary connected CW complexes. By attaching cells to kill off higher homotopy groups, we obtain inclusions $X \to K(\pi_1 X,1)$ and $Y \to K(\pi_1 Y, 1)$ inducing isomorphisms on $\pi_1$. We've already observed that we can fill in a map between $K(\pi_1 X, 1)$ and $K(\pi_1 Y, 1)$ realizing any desired homomorphism, so we have: $$\begin{array}{ccc} X & & Y \\ \downarrow & & \downarrow \\ K(\pi_1 X, 1) & \rightarrow & K(\pi_1 Y, 1) \end{array}$$ We want to know whether the top edge can be filled in making the diagram commute.

Up to homotopy, this can be done when the map $$X \to K(\pi_1 X, 1) \to \operatorname{cofib}(Y \to K(\pi_1 Y, 1))$$ is nullhomotopic. This gives a sufficient (but not necessary) condition for an affirmative answer to your question. I believe this also recovers the condition in Balarka's comment.

JHF
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