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Consider the set $Map(T^4,S^2)$ of continuous maps from the 4 dimensonal torus $T^4$ to the 2 dimensional sphere $S^2$, endowed with compact-open topology, can we show it is not connected? How can we calculate its singular homology and $\pi_1$?

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    In spite of being more or less standard notations, it'd be better for your question's clearity if you'd define $;T^4,,,S^2;$ ... – DonAntonio Nov 05 '13 at 15:24

2 Answers2

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For the first part

Hint 1: $$Map(X\times Y,Z)\cong Map(X,Map(Y,Z))$$

Hint 2: $$\pi_i(Map(S^1,X))\cong\pi_{i+1}(X)$$

Hint 3: $$\pi_4(S^2)\cong \mathbb{Z}_2$$

For the second and third parts

Hint 4: $$\pi_5(S^2)\cong\mathbb{Z}_2$$

Hint 5: $$H_1(X)\cong \pi_1(X)^{ab}$$

Hint 6: For higher $H_k$, I think you'll need to iterate the Leray spectral sequence as far as I can tell, which will be messy - there may be an easier way which can be applied to the sphere and its loop-spaces (see this question).

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Dan Rust
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  • Feel free to wait to or not answer this since it would provide more than just hints, but is there a good way to give a geometric description of these nontrivial homotopies? – Zach L. Nov 05 '13 at 16:33
  • @ZachL. Which ones? The first hint is a common homeomorphism/isomorphism colloquially known as currying which is valid in any cartesian closed category - in this case compactly generated Hausdorff spaces. The second is a result of the long exact sequence of the loop-space fibration $\Omega(X)\to P(X)\to X$ where $P(X)$ is the path-space of $X$ and $\Omega(X)\simeq Map(S^1,X)$. – Dan Rust Nov 05 '13 at 17:02
  • Sorry that I am not clear with notations about path spaces, What do you mean by $P(X)$? Does it mean the space of all paths that joins any two points in $X$?Is there any reference for that fibration? I didn't find it in wiki.. –  Nov 06 '13 at 07:11
  • @mqx You should be able to find the result in any decent book on algebraic topology - for instance Hatcher proves $\pi_i(Map(S^1,X))\cong\pi_{i+1}(X)$ in several different ways (including the above fibration method). Just search for 'loop space' in his text. – Dan Rust Nov 06 '13 at 10:15
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    Hint 2 is not true for the space of unbased maps from $S^1$ to a space $X$. You get the free loop space, not the based loop space, and in general their homotopy groups differ in every degree. – Qiaochu Yuan Dec 15 '14 at 07:29
  • @QiaochuYuan thanks for this. I remember making a similar mistake recently but thankfully catching it that time. If you make your answer self contained at some point, I can delete this answer - just ping me. – Dan Rust Dec 15 '14 at 09:49
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The accepted answer is incorrect. The problem is in Hint 2, which conflates based maps with unbased maps, and in particular which conflates the based loop space $\Omega X$ of a pointed space $(X, x)$ (the space of maps $S^1 \to X$ sending a fixed basepoint in $S^1$ to $x$) with the unbased or free loop space $LX$ of a space $X$ (the space of maps $S^1 \to X$, with no further hypotheses). Hint 1 and Hint 2 together were supposed to convince you that the space you're looknig at is the 4-fold based loop space of $S^2$, which satisfies

$$\pi_0(\Omega^4 S^2) \cong \pi_4(S^2) \cong \mathbb{Z}_2$$

but that's not true; the 4-fold based loop space of $S^2$ is the space of maps $S^4 \to S^2$ sending a fixed basepoint of $S^4$ to a fixed basepoint of $S^2$, and has nothing to do with $T^4$. The space you're looking at is in fact the 4-fold free loop space $L^4 S^2$.

Qiaochu Yuan
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    We re looking at the space of maps $T^2\times T^2\to S^2$, and to such a thing we can attach a bidegree, which will be homotopy invariant. If that works out, we can just contruct two maps with different bidegree (one of bidegree $(1,0)$ and the obvious swap should do the trick) aand that'll show it is not connected. – Mariano Suárez-Álvarez Dec 15 '14 at 08:05
  • Right. More generally I think I know how to compute the homotopy groups of $LX$ from the homotopy groups of $X$, but applying this procedure $4$ times gets a little messy and I don't know a good way of organizing the results. – Qiaochu Yuan Dec 15 '14 at 19:32
  • Well, unless you plan to do it rationally, that is going to be very difficult. Rationally, using minimal models and rational homotopy theory, it's a standard computation. – Mariano Suárez-Álvarez Dec 16 '14 at 03:21
  • We have $Map(T^4,S^2) \simeq Map(T^2,Map(T^2,S^2))$ and it is therefore enough to show that $Map(T^2, S^2)$ is disconnected since then the space of maps from $T^2$ to a disconnected space is also disconnected. The last assertion is clear by looking at $H_2$. Namely, we can construct both a map of degree 1 and a map of degree -1 and they are not homotopic, since they disagree on homology. – KotelKanim Jul 19 '16 at 10:56