I have to prove the following:

Using the canonical isomorphism $$\pi_1(S^1\times S^1,(1,1)) \cong \pi(S^1,1)\times \pi_1(S^1,1),$$ show that every endomorphism of the group $\pi_1(S^1\times S^1,(1,1))$ can be expressed as $f_\ast$ for some continuous map $f : S^1\times S^1 \to S^1\times S^1.$

[Note: We already know that the last property is true for $\pi_1(S^1,1)$, and it can be used.]

My attempt was trying to write the homomorphism as a direct product of homomorphisms from $\pi_1(S^1,1)$ to itself in order to use the Note. I can't conclude the solution since I think I can't recover the original homomorphism after considering the homomorphisms induced in each factor $\pi_1(S^1,1).$

I hope someone could help me and give me the idea to conclude the problem. It's important that it's my first course on algebraic topology and I'm only given the tools of fundamental groups. Thanks in advance!

  • $\begingroup$ Thanks for your comments @freakish. The first comment is a direct road to the solution. I'm looking for a solution that uses the property for $S^1.$ In your second comment I think that you solve my difficulties, I was considering only the inclusion and projection always in the same factor. I was forgetting some combinatorics intrinsic to the problem of reconstruct a map. I think that you must post your comments as an answer. $\endgroup$ – DrinkingDonuts Jun 19 at 13:28

First of all by $\mathbb{Z}^n$ I will understand any free abelian group of rank $n$. In particular I will literally write $\pi_1(S^1)=\mathbb{Z}$. This is to avoid many details arising from considering isomorphisms. Although a careful reader should try to fill the missing details.


Consider $\psi:\mathbb{Z}^n\to\mathbb{Z}^n$ a homomorphism, $J_k:\mathbb{Z}\to\mathbb{Z}^n$, $J_k(x)=(0,\ldots,0,x,0,\ldots,0)$ the standard group embedding onto $k$-th coordinate and $P_k:\mathbb{Z}^n\to\mathbb{Z}$ the standard projection of $k$-th coordinate. So for any $1\leq s,t\leq n$ we have an induced homomorphism

$$\psi_{st}:\mathbb{Z}\to\mathbb{Z}$$ $$\psi_{st}=P_s\circ \psi\circ J_t$$

So any endomorphism $\psi:\mathbb{Z}^n\to\mathbb{Z}^n$ induces a set of homomorphism. The main observation is that this works in the other direction as well. If $\{\psi_{st}:\mathbb{Z}\to\mathbb{Z}\}$ is a set of homomorphisms then we can recreate $\psi$ via

$$\psi(x_1,\ldots,x_n)=\left[\begin{matrix} \psi_{11} & \psi_{12} & \cdots & \psi_{1n} \\ \psi_{21} & \psi_{22} & \cdots & \psi_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ \psi_{n1} & \psi_{n2} & \cdots & \psi_{nn} \\ \end{matrix}\right]\left[\begin{matrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{matrix}\right]$$

So it's a matrix multiplication where $\psi_{st} x$ should be understood as $\psi_{st}(x)$.

And these two operations, i.e. $\psi\mapsto\{\psi_{st}\}$ and $\{\psi_{st}\}\mapsto\psi$ are inverses of each other. Meaning $End(\mathbb{Z}^n)$ is equinumerous (even ring isomorphic) with $\mathbb{M}_{n^2}(End(\mathbb{Z}))$. In particular every set of morphisms $\{\psi_{st}\}$ uniquely determines the induced morphism $\psi$.

Side note: here $\mathbb{Z}$ didn't play any significant role. And indeed, this construction works well for any abelian group $A$. It even generalizes to modules. But it doesn't work for non-abelian groups.

Algebraic Topology

Throughout $1\in S^1\subseteq\mathbb{C}$ will be the implicit base point for all pointed spaces.

Back to the problem. So we know that $\pi_1((S^1)^n))=\mathbb{Z}^n$ is free abelian of rank $n$ and so any endomorphism is uniquely determined by $n^2$ corresponding $\mathbb{Z}\to\mathbb{Z}$ morphisms.

Consider continuous embedding $j_k:S^1\to (S^1)^n$, $j_k(z)=(1,\ldots,1,z,1,\ldots, 1)$ and continous standard projections $p_k:(S^1)^n\to S^1$. All you have to realize now is that $j_{k*}=J_k$ and $p_{k*}=P_k$. This requires some effort (but is not very hard) and I will not deal with it here.

But if you already know that then the rest becomes simple. You take any $\psi:\mathbb{Z}^n\to\mathbb{Z}^n$, you consider $\psi_{st}:\mathbb{Z}\to\mathbb{Z}$ as earlier, then you take $f_{st}:S^1\to S^1$ such that $f_{st*}=\psi_{st}$ and finally compose it back to $f:(S^1)^n\to(S^1)^n$ map in such a way that $f_{st}=p_s\circ f\circ j_t$. You do that by definining

$$f(x_1,\ldots,x_n)_k=f_{k1}(x_1)\cdots f_{kn}(x_n)$$

with complex multiplication on the right side. So it's similar to the matrix multiplication as defined in Algebra section, except we use complex number multiplication instead of the standard integer addition.

Side note: $f_{st}(1)=1$ because we work with pointed spaces.

The rest follows from the functoriality of $\pi_1$:

$$P_s\circ\pi_1(f)\circ J_t=\pi_1(p_s)\circ\pi_1(f)\circ\pi_1(j_t)=\pi_1(p_s\circ f\circ j_t)=\pi_1(f_{st})=\psi_{st}$$

Since both $\pi_1(f)$ and $\psi$ are determined by the same set of morphisms $\{\psi_{st}\}$ then they have to be equal.


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