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Could you help solve this?

Let $f: X \rightarrow Y$ be a function and $Y = A \underline \cup B$ where $\underline \cup$ means the disjoint union.

Show that:

a) $f^{-1}(A) \underline \cup f^{-1}(B) = X \\ \\ $

b) $f|_{f^{-1}(A)}$ is an injection and $f|_{f^{-1}(B)}$ is an injection then $f$ is an injection. and $f|_{f^{-1}(A)}$ is a restricted function.

The main problem is that I'm not sure how to use the disjont sum. I'd really appreciate a thorough explanation. Thank you.

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2  
I don't see any disjoint sum. I see a disjoint union, but that just means $A$ and $B$ are disjoint sets. –  Gerry Myerson Jan 6 '13 at 15:09
    
$Y = A \underline \cup B$ just means that $Y = A \cup B$ and that $A$ and $B$ are disjoint, i.e. $A \cap B = \varnothing$. Have a think about why (a) is true; it will give you clues as to how to solve (b). –  Clive Newstead Jan 6 '13 at 15:11

1 Answer 1

up vote 2 down vote accepted

(a) It’s clear that $X=f^{-1}[A]\cup f^{-1}[B]$, so all that remains is to show that this is a disjoint union, i.e., that $f^{-1}[A]\cap f^{-1}[B]=\varnothing$. Suppose that $x\in f^{-1}[A]\cap f^{-1}[B]$; then $f(x)\in A$ and $f(x)\in B$, so ... ?

(b) Let $X_A=f^{-1}[A]$ and $X_B=f^{-1}[B]$, and suppose that $f\upharpoonright X_A$ and $f\upharpoonright X_B$ are injections. To show that $f$ is an injection, suppose that $x_0,x_1\in X$ and $f(x_0)=f(x_1)$; you need to show that $x_0=x_1$. Just consider the possible cases:

  1. $x_0,x_1\in X_A$: then $f(x_0)=(f\upharpoonright X_A)(x_0)$ and $f(x_1)=(f\upharpoonright X_A)(x_1)$, and $f\upharpoonright X_A$ is injective, so ... ?
  2. $x_0,x_1\in X_B$: this is clearly just like the previous case.
  3. $x_0\in X_A,x_1\in X_B$: then $f(x_0)\in A$ and $f(x_1)\in B$. Is it possible that $f(x_0)=f(x_1)$?
  4. $x_0\in X_B,x_1\in X_A$: this is clearly just like the previous case.
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