Is Map($T^4$,$S^2$) connected? 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 $\pi_1$?
 A: 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).
A: 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 $L^4 S^2$.
