Prove that is $A$ is a subspace of $X$, the inclusion function $j: A \to X$ is continuous

Question

Prove that is $A$ is a subspace of $X$, the inclusion function $j: A \to X$ is continuous

There is a proof of this in Munkres : Topology a First Course that I don't understand.

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Munkres' Proof

If $U$ is open in $X$, then $j^{-1}(U) = A \cap U$, which is open in $A$ by definition of the subspace topology.

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What I can't understand is why $j^{-1}(U) = A \cap U$. I understand that the subspace topology $\mathcal{T}_A = \{ A \cap U \ | \ U \in \mathcal{T}_X\}$, where $\mathcal{T}_X$ is the topology on $X$.

Is it simply because $j$ maps $A$ into $X$, that the inverse is defined as $j^{-1}(U) = \{x \in A \ | \ j(x) \in U\}$ and since both $A \subset X$ and $U \subset X$, we can define $j^{-1}(U) = A \cap U$?

Answer 1

By definition, $$j^{-1}(U)=\{x\in A\mid j(x)\in U\}.$$ But $j(x)=x$ for all $x$, so this is the same as $$\{x\in A\mid x\in U\}.$$ That is, $x\in j^{-1}(U)$ iff $x\in A$ and $x\in U$. That is, $x\in j^{-1}(U)$ iff $x\in A\cap U$.