# Doubt in the definition of manifolds?

Let $$M$$ be a Hausdorff space. Assume that there exists an open covering $$\{U_{\alpha} : \alpha \in A\}$$ of $$M$$ and homeomorphisms $$\varphi_{\alpha}$$, from $$U_{\alpha}$$ onto an open subset $$\varphi_{\alpha}(U_{\alpha})$$ of $$\mathbb{R}^m(\alpha)$$ with $$m(\alpha)$$ a nonnegative integer.

There exists $$k \in \mathbb{N} \cup \infty$$ such that whenever for $$\alpha,\beta \in A$$ we have then the map $$U_{\alpha} \cap U_{\beta} \ne \phi$$ then the map: $$\varphi_{\alpha} \circ \varphi_{\beta}^{-1}:\varphi_{\beta}(U_{\alpha} \cap U_{\beta}) \to \varphi_{\alpha}(U_{\alpha} \cap U_{\beta})$$ is $$C^k$$.

The intuition of the definition behind the manifold is that it behaves locally as a subset of $$\mathbb{R}^n$$ ( there exists a homeomorphism between $$U_p$$ a neihbourhood of all possible points $$p \in M$$ such that $$\varphi(U_p)$$ is a subset of $$\mathbb{R}^n$$) such that any function $$f:U \to \mathbb{R}$$ is $$C^k$$ [($$f \circ \varphi^{-1}):\varphi(U) \to \mathbb{R}$$ is $$C^k$$($$k$$ times differentiable)].

I can understand the first line of the definition but I am stuck with $$\varphi_{\alpha} \circ \varphi_{\beta}^{-1}:\varphi_{\beta}(U_{\alpha} \cap U_{\beta}) \to \varphi_{\alpha}(U_{\alpha} \cap U_{\beta}) is C^k$$ - how is it related?

• You should probably search for better definitions or a book. John Lee's book or Loring Tu are good ones. Here is a reference math.stackexchange.com/questions/1897936/… . Your second para does not make any sense to me and is incomplete. Also it should be $\phi(U_{\alpha}\cap U_{\beta})$ Mar 8, 2022 at 15:30
• $\varphi_\beta$ isn't defined on $U_\alpha .$ It looks to me as if this was a typo: '\cup' instead of '\cap'. Mar 8, 2022 at 15:32
• Yes I just changed it. Mar 8, 2022 at 15:33
• @Mr.GandalfSauron I went through the link but i really can't make out anything from that. My problem is that the intuition behind $\varphi_{\alpha} \circ \varphi_{\beta}^{-1}$ is $C^k$? Mar 8, 2022 at 15:38

Okay. To understand the intuition you need to have the notion of smooth maps from a manifold to another. Notice that you do not(yet) have the notion of differentiability on the manifold but you do have on $$\mathbb{R}^{n}$$. So to define "smoothness" you have to sort of bring it to $$\mathbb{R}^{n}$$.
Now a map $$F:M\to N$$ is smooth when for a point $$p$$ and charts $$(U,\phi)$$ of $$p$$ and $$(V,\psi)$$ of $$F(p)$$ , $$\psi\circ F\circ \phi^{-1}$$ is a smooth map from $$\phi(U)\subset \mathbb{R}^{m}\to \mathbb{R}^{n}$$ . $$m$$ and $$n$$ are dimensions of $$M$$ and $$N$$.
• identity map is smooth as it is a composition of $\phi \circ \phi^{-1}$ and the compoistion of differentiable maps are differentiable? Mar 8, 2022 at 15:53
• Well yes and no. How about for different charts $(ϕ_α,U_α)$ and $(ϕ_β,U_β)$ for the same point p?. In that case the the identity map is smooth if $ϕ_α∘ϕ^{−1}_β$ and $ϕ_β∘ϕ^{−1}_α$ is a smooth map from a subset of $\mathbb{R}^{n}$ to another subset. The definition makes sure that when talking about smoothness, you are not limited by choice of charts. Mar 8, 2022 at 16:11