# Help with a proof involving images and intersections

• November 29th 2010, 01:33 AM
Lprdgecko
Help with a proof involving images and intersections
I have no idea where to even start on this proof. Any help would be greatly appreciated :) I think one of the problems I am having is that I don't quite understand the concept of an image. I know that, to show it is injective, I will have to show that h(x) = h(y) but I don't know how to start...

(I apologize if this is difficult to read)

Let h: X --> Y be a function. Show that h is injective iff im(h)(A intersect B) = im(h)(A) intersect im(h)(B) for all A,B subsets of X.

**** im(h)(A) means the image of h.... written as an h with a sub-star. Not sure if the notation I typed is correct.
• November 29th 2010, 02:22 AM
FernandoRevilla
For example:

$\Rightarrow\;)$ . Suppose $h$ is injective. Then:

(i) $h(A\cap B)\subset h(A)\cap h(B)$ is always true (being $h$ injective or not).

(ii) $y\in h(A)\cap h(B)\Rightarrow \begin{Bmatrix} y\in h(A)\\y\in h(B)\end{matrix}\Rightarrow \begin{Bmatrix} y=h(a):a\in A\\ y=h(b):b\in B\end{matrix}$

But $y=h(a)=h(b)\Rightarrow a=b \Rightarrow a\in A\cap B\Rightarrow y=f(a)\in h(A\cap B)$ .

Try someting for $\Leftarrow\;)$

Regards.

Fernando Revilla
• November 29th 2010, 02:52 AM
emakarov
Quote:

I know that, to show it is injective, I will have to show that h(x) = h(y) but I don't know how to start...
You should start by reviewing the definition of injectivity. A function h is called injective if for every x and y from the domain, h(x) = h(y) implies x = y. So you don't show that h(x) = h(y); you assume it. You should also review examples and proofs involving injectivity from your textbook or lecture notes to develop some intuition about it.

It may be helpful to visualize the situation by looking at the graph of h. Consider, for example, h(x) = sin(x) on $[0,\pi]$. Let $A = [0,\pi/2]$ and $B = (\pi/2,\pi]$. Then $A\cap B=\emptyset$, but h[A] = h[B] = [0, 1], so $h[A]\cap h[B]=[0,1]$. (I'll write h[A] for im(h)(A).) In fact, $h[A\cap B]\subseteq h[A]\cap h[B]$ holds for every function, not necessarily injective. The converse inclusion does not hold here because h is very non-injective: for every $x\in[0,\pi/2), h(x)=h(\pi-x)$. If $A\cap B=\emptyset$, as it is here, the requirement that $h[A]\cap h[B]\subseteq\emptyset$ means that h assumes different values on non-overlapping segments A and B. This is basically injectivity: mapping distinct arguments into distinct results.

Formally, one has to prove both directions of iff: if h is injective, then $h[A\cap B]=h[A]\cap h[B]$ for all A, B and vice versa. For the forward direction, assume that h is injective and fix some A and B. The equality of $h[A\cap B]$ and $h[A]\cap h[B]$ is also proved in a standard way by showing $h[A\cap B]\subseteq h[A]\cap h[B]$ and $h[A\cap B]\supseteq h[A]\cap h[B]$. For each inclusion, you assume that some y is in the smaller set and show that it is in a bigger one.

Fernando showed a hint for the forward direction. ( $\subset$ in (i) should be understood as nonstrict inclusion.)