# Intersection of Images of a function

I'm trying to understand intuitively why the image ( under some function ) of the intersection of subsets of the domain of that function is only contained ( and not equal ) to the intersection of the images of the corresponding sets.

What's the intuition on to why $f\left(\bigcap_{i\in I}A_i\right)\subseteq\bigcap_{i\in I}f(A_i)$ and not $f\left(\bigcap_{i\in I}A_i\right)=\bigcap_{i\in I}f(A_i)$.

Whats the intuition on to why the intersection of images might contain more elements than the image of intersections ?
I even have one counter-example, suppose $f :\mathbb R \to\mathbb R$ defined as $f(x) = x^2$. Then indeed, if A1 = [-1,0] and A2 = [0,1], then the image of the intersections must be contained ( and not be equal ) to the intersection of images.

But i'm trying to abstract and see what's the general requirement for the function and for the family of sets so that the image of intersection is contained and not equal to the intersection of images.

• I think there are at least two or three or seven hundred threads with proofs about this fact, discussions and otherwise intuition. Have you read any of them? – Asaf Karagila Aug 18 '14 at 22:31
• Because functions do not need to be one-to-one. – Lee Mosher Aug 18 '14 at 22:31
• Putting it very loosely, $f$ can glue stuff together. Or: things can meet in the image that don't meet in the domain. To see this, you can very easily make up a function with the property that there are (say) three disjoint sets in the domain whose images coincide (say, they are the nonempty set $K$). That is, all the $A_i$ are disjoint (so $f(\cap_iA_i)=f(\varnothing)=\varnothing$), but $\cap_i f(A_i)=\cap_i K=K$. – MPW Aug 18 '14 at 22:52
• MPW, i see. So, like Lee Mosher pointed out, is it true that if f is an injection then that proposition holds specifically for the equality ( instead of the containment ), regardless of the Subsets Ai we choose of the domain ? Also, could some choice of a non-injective function and specific subsets of its domain result in the image of intersections of those subsets as being equal to ( and not only contained in ) the intersection of images of those subsets ? Thanks in advance – nerdy Aug 19 '14 at 5:40

Intuitively this is because a mapping $f:A\rightarrow B$ may identify elements in the sense of:

$$f(x)=f(y) \text{ in } B \text{ where } x\neq y \text{ in }A$$

but a mapping by definition may not distinguish or separate an element into two in this sense:

$$f(x)\neq f(y)\text{ in } B \text{ where } x=y \text{ in } A$$

Say $A_1, A_2\subset A$. Think of $f(A_1)\cap f(A_2)$ as the "identifications" done by $f$ in $B$ of elements from $A$.

Then the mapping $f$ necessarily "identifies" all elements from $A_1\cap A_2$ - i.e.: $$f(A_1\cap A_2)\subset f(A_1)\cap f(A_2)$$

All that can happen additionally during $f$ mapping is that there are more points from $A_1$ and $A_2$ that are not from $A_1\cap A_2$ that are identified by $f$ in $B$ in which case we have $$f(A_1\cap A_2)\not\supset f(A_1)\cap f(A_2)$$

Let me give you an example:

$\text{Let }X = \{1, 2\} \text{ and } f(1) = f(2) = 1$.

$\implies f[\{1\}\cap\{2\}]=\varnothing \text{ and }f[\{1\}]\cap f[\{2\}]=\{1\}$

$\implies f[\{1\}\cap\{2\}] \neq f[\{1\}]\cap f[\{2\}]$