TITLE: Degree 1 maps from $\mathbb S^n$
QUESTION [8 upvotes]: Suppose that $f:S^n \to M$ is a map from the $n$-sphere to a simply-connected $n$-dimensional manifold that induces an isomorphism on top homology. I wonder if it's true that $f$ is already a homotopy equivalence.
I will first present the arguments of a fellow user which show that $M$ can have no non-torsion elements in reduced homology below dimension $n$. Assuming the contrary, there exists a primitive non-torsion element  $\alpha \in H^k(M,\mathbb Z)$ for $0 < k < n$. By Poincaré-duality, there exists some $\beta \in H^{n-k}(M,\mathbb Z)$ such that $\alpha \cup \beta$ is generator of $H^n(M,\mathbb Z)$. But $f^*$ is natural with respect to cup products and an isomorphism in top cohomology. We arrive at a contradiction, since $H^k(S^n) = 0$ for $0 < k < n$.
Taking $\mathbb Q$ coefficients (thereby killing torsion), we have that $f$ induces an isomorphism on all homology groups with $\mathbb Q$ coefficients. Since $M'$ was assumed to be simply connected, it follows by the rational Hurewicz theorem that $f$ induces an isomorphism on all homotopy groups tensored with $\mathbb Q$. The Whitehead theorem, however, cannot be imediately applied now. Is there any other way to conclude that $f$ is a homotopy equivalence ?

REPLY [7 votes]: While the question as asked has already been resolved, let me remark that the hypothesis that $M$ is simply connected is unnecessary.  Indeed, suppose $M$ is possibly not simply connected, and let $p:\tilde{M}\to M$ be its universal cover.  We may assume $n>1$, so that our degree $1$ map $f:S^n\to M$ lifts to a map $\tilde{f}:S^n\to\tilde{M}$.  It follows that the isomorphism $f_*:H_n(S^n,\mathbb{Z})\to H_n(M,\mathbb{Z})$ factors through $H_n(\tilde{M},\mathbb{Z})$.  In particular, this means that $H_n(\tilde{M},\mathbb{Z})\neq 0$ so $\tilde{M}$ must be compact; it follows that $p$ is a finite cover.  The degree of $p$ (i.e., the map induced on $H_n$) is then equal to the degree of $p$ as a covering map.  Since $1=\deg f=\deg \tilde{f}\deg p$, we must have $\deg p=1$.  Thus $p$ is a degree $1$ cover and hence a homeomorphism, so $M\cong \tilde{M}$ is simply connected.


TITLE: $(-3)^{3/2} \neq (-3)^{6/4}$
QUESTION [1 upvotes]: $(-3)^{\frac{3}{2}}=-3\sqrt{3}i$
$(-3)^{\frac{6}{4}}=\sqrt{27}$
(not the same thing).
What's the deal? It's interesting because people work with fractional exponents all the time and I've never seen someone bother to check whether the top and bottom maintain their parity when canceling, but clearly it makes a difference if you can have a negative base.
More precisely, how are exponents (especially of negative numbers) defined (in a rigorous sense), so that I can understand the problem here?
I don't think the solution is just to form a convention in which you simplify as much as possible before doing operations. I know it would give consistent results, but by the same reasoning, we could have chosen 0!=0. We chose not to make it that way for good reason. There are many applications in which 0! = 1 is the only elegant possibility. Having asked that... does anyone know of applications of this sort of thing?

REPLY [1 votes]: You have to be careful with these kind of things if your base is not a non-negative real number. For example,
$$1=1^{1/2}=[(-1)^2]^{1/2}=-1.$$
The reason for this can be found when you look at the "true" definition of $x\mapsto a^x$ when $a\in\mathbb{C}\setminus[0,\infty)$. We define this as:
$$a^x:=e^{x\log a}$$
which of course requires some kind of definition for the logarithm. The usual one is
$$\log x:=\log|x|+i\arg x$$
which is of course multi-valued since $\arg$ is. We thus don't have, in general, $(ab)^c=a^cb^c$ or $(a^b)^c=a^{bc}$.