Smooth map is diffeomorphism iff its pullback is an algebra isomorphism?

Question

newbie here in Diff Geo/Man!

I am given topological spaces $M$ and $N$, their algebras of continuous functions, $C(M), C(N)$, respectively, and a map $F^* = — \circ F : C(N) \to C(M)$, which is easily seen to be an algebra homomorphism. Also easy to see is that $F$ is smooth iff $F^*[C^\infty(N)] \subseteq C^\infty(M)$.

Also easily verifiable is that íf $F$ is a diffeomorphism, then $F^*$ is an algebra isomorphism. The converse, however, keeps me awake. Probably because I am so close.

I attempted to focus only on (smooth) constant functions, say, $$\text{Const}(M) := \Gamma(M) := \{c_m : M \to \mathbb R \mid m \in M, \text{im}(c_m) = \{m\}\} \subset C^\infty(M)$$ (And similarly for $N$, of course.)

I also know/believe, is that íf an inverse to $F$ exists, we must have $(F^*)^{-1} = — \circ F^{-1}$. The problem is I haven't been able to construct $F^{-1}$...

So, for some harmless diagrams:

$\require{AMScd}$ \begin{CD} M @>{\mu_0}>> \text{Const}(M) @>{\iota_M|_{\Gamma(M)}}>> C^\infty(M) \\ @V{F}VV @AA{F^*|_{\Gamma(N)}}A @AA{F^*}A\\ N @>{\nu_0}>> \text{Const}(N) @>{\iota_N|_{\Gamma(N)}}>> C^\infty(N) \end{CD}

(We may not even need the right square.) As $\mu_0 : M \to \Gamma(M), \, m_0 \mapsto (c_{m_0} : m \mapsto m₀)$ is in clear bijection with $M$, similar for $\nu_0$, I went out to construct the inverse $G$ of $F$ as follows:

$$G := \mu_0^{-1} \circ F^*|_{\Gamma(N)} \circ \nu_0 : N \to M. $$

However, when I perform my calculation, even while realising that precomposing a constant map with any map, remains that constant map ($c_{F(m)} \circ F = c_{F(m)}$), I just get the wrong answer: let $m \in M$. Then

\begin{align*} G \circ F(m) &= ((\mu_0^{-1} \circ F^*|_{\Gamma(N)} \circ \nu_0)\circ F)(m) = (\mu_0^{-1} \circ F^*|_{\Gamma(N)}) \circ \nu_0 (F(m)) \\ &=(\mu_0^{-1} \circ F^*|_{\Gamma(N)})(c_{F(m)}) = \mu_0^{-1}(c_{F(m)} \circ F) \\ &= \mu_0^{-1}(c_{F(m)}) = F(m) \neq m = \text{id}(m). \end{align*}

Can anyone spot a mistake? It looked clear enough to me that we wouldn't have to incorporate the entire diagram, but perhaps I should have, and instead have worked with $(\iota_M|_{\Gamma(M)} \circ \mu_0)^{-1}$, but I didn't think it necessary.

Any input greatly appreciated! Cheers!

Answer 1

Instead of trying to get the inverse to just "pop out", I would show that $F$ is a bijection. First, $F$ is injective since $F^*$ is surjective (if $F(x)=F(y)$ and $x\neq y$, then $F^*$ would not be surjective since its image would not contain any function that takes different values on $x$ and $y$). Then to see that $F$ is surjective, let $x\in N$ and take $f\in C^\infty(N)$ that vanishes only at $x$. Then $f$ is not a unit in the ring $C^\infty(N)$, so $F^*(f)$ is not a unit in the ring $C^\infty(M)$. This means $f\circ F$ must vanish somewhere, which means exactly that $x$ must be in the image of $F$.

So, $F$ is a bijection, and now you can conclude that $F^{-1}$ is smooth since composition with it sends smooth functions to smooth functions (since composition with $F^{-1}$ is just given by $(F^*)^{-1}$).

(Alternatively, for something more along the lines of what you were trying, you can "extract" the points of $M$ from $C^\infty(M)$ by identifying $m\in M$ with the homomorphism $C^\infty(M)\to\mathbb{R}$ that takes a function and evaluates it at $m$. This gives a stronger result, that any ring isomorphism $C^\infty(N)\to C^\infty(M)$ (which is not assumed to be given by composition with a map $M\to N$) is given by composition with a diffeomorphism $M\to N$. See the answers to Can we recover a compact smooth manifold from its ring of smooth functions? for more details on this.)