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The premise is: Let f: X $\to$ Y be a continuous map, and A $\subset$ X and B $\subset$ Y. Assume f(A) $\subset$ B; then we can define a restriction map g: A $\to$ B by g(a)=f(a)

a). For any V$\subset$Y, show that g$^{-1}$(V $\cap$ B)=f$^{-1}$(V) $\cap$ A

I'm not entirely sure where to begin but I believe I need to show that f is a surjective map and thus I can continue with f$^{-1}$(V) = g$^{-1}$(V) if V $\subset$ B

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  • $\begingroup$ This has nothing to do with topology, it is only a question about sets and functions. You can prove that the identity holds by proving the two inclusions separately (i.e. for $\subseteq$, pick a $x\in A$ such that $g(x)\in V\cap B$ and prove that $x$ is in the right hand side). $\endgroup$ – zarathustra Feb 27 '17 at 21:53
  • $\begingroup$ Thank you, that's what I thought but this is for my topology class so I thought I was missing something. $\endgroup$ – Niko L Feb 27 '17 at 22:01

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