Does Injectivity Ensure f(Aᶜ) is a Subset of f(A)ᶜ?

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Homework Statement


Let f: X \rightarrow Y be a function. Suppose A is a subset of X. Show that if f is injective, then f(A^{c})\subseteq f(A)^{c}.


Homework Equations





The Attempt at a Solution


If x \in A^{c}, then there is a y \in f(A^{c}) such that f(x)=y. As f is an injection, y \notin f(A), hence y \in f(A)^{c}.

Is that alright?
 
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bedi said:

Homework Statement


Let f: X \rightarrow Y be a function. Suppose A is a subset of X. Show that if f is injective, then f(A^{c})\subseteq f(A)^{c}.


Homework Equations





The Attempt at a Solution


If x \in A^{c}, then there is a y \in f(A^{c}) such that f(x)=y. As f is an injection, y \notin f(A), hence y \in f(A)^{c}.

Is that alright?
You're not using the correct definition of one-to-oneness. For an injection, given a y value, there is exactly one x value. The usual example for a function that isn't one-to-one is f(x) = x2. Here, both 2 and -2 map to 4.

It seems to me that what you are doing is pairing one number in the domain (x) with two numbers in the codomain (y1 and y2), where y1 ##\in## f(A) and y2 ##\in## f(AC). That isn't even a function, let alone an injective function.

What you need to show is that if y ##\in## f(AC), then it follows that y ##\in## [f(A)]C. I think that's what you mean by f(A)C.
 
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