Smith, J. (2018). Introduction to Differential Manifolds. Oxford University Press.

Brown, A. (2019). Differential Manifolds and Their Applications. Wiley.

Miller, R. (2020). Stochastic Methods in Differential Manifolds. Springer.

Johnson, L. (2021). Applications of Differential Manifolds in Physics. Cambridge University Press.

Chen, H. (2022). Advanced Topics in Differential Manifolds: Fiber Bundles and Chern Classes. MIT Press.

Problem 1.1. Show that the induced topology indeed satisfies the definition of a topology.

Solution:
Let $\mathcal{U}=\left{U_i\right}$ be the topology of $Y$ and $X \subset Y$. The induced topology is $\mathcal{V}=$ $\left{U_i \cap X \mid U_i \in \mathcal{U}\right}$
i) $\emptyset=\emptyset \cap X$ so since $\emptyset \in \mathcal{U}$ we have $\emptyset \in \mathcal{V}$.
ii) $X=X \cap Y$ so since $Y \in \mathcal{U}$ we have $X \in \mathcal{V}$
iii) Take $V_i=U_i \cap X \in \mathcal{V}$ then
\begin{aligned} V_1 \cap V_2 \cap \ldots \cap V_n & =\left(U_1 \cap X\right) \cap\left(U_2 \cap X\right) \cap \ldots \cap\left(U_n \cap X\right) \ & =\left(U_1 \cap U_2 \cap \ldots \cap U_n\right) \cap X \ & \in \mathcal{V} \end{aligned}
iv) For an arbitrary number of $V_i$ ‘s:
$$\bigcup_i V_i=\bigcup_i\left(U_i \cap X\right)=\left(\bigcup_i U_i\right) \cap X \in \mathcal{V}$$

Problem 1.2. Why aren’t closed subsets of $\mathbb{R}^n$, e.g. a disk with boundary or a line in $\mathbb{R}^2$, along with the identity map charts (note that in its own induced topology any subset of $\mathbb{R}^n$ is an open set)?

Solution: If $C \subset \mathbb{R}^n$ is closed we may still view it as an open set in its own induced topology. The identity map $i d: C \rightarrow \mathbb{R}^n$ is certainly continuous (in inverse image of an open set is open by definition of the induced topology). It is also clearly $1-1$ and hence is a bijection onto its image. However consider $i d^{-1}: i d(C) \subset \mathbb{R}^n \rightarrow C$ this has
$$\left(i d^{-1}\right)^{-1}(C)=C$$
but since $C$ is open in its own induced topology and closed in the topology of $\mathbb{R}^n$ we see that $i d^{-1}$ is not continuous (as $\left(i d^{-1}\right)^{-1}$ of an open set is closed). Thus $i d$ is not a homeomorphism.

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