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Does this sound kind of correct or at least am I on the right track here?
Lets say: some x = f(x) for some a for which a ∈ S ∩ T, so x ∈ f(S∩T). And for some y = f(y) for some b for which b ∈ S and b ∈ T, so y ∈ f(S) ∩ f(T). And because f is 1-1 f(x) = f(y) or x=y which would imply f(S∩T) = f(S) ∩ f(T)
Therefore f(S∩T)=f(S)∩f(T)
I feel like I get lost once brining the 1-1 part in, and just in general doing it, but I also feel like I am on track I'm just not sure if its correct or not. Any advice would be appreciated.

Amber
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  • Welcome to Math.SE. Posts here are formatted in MathJax. Here's a helpful guide: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – pyridoxal_trigeminus Nov 06 '21 at 23:49
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    Proving $f(S \cap T) \subseteq f(S) \cap f(T)$ should be pretty straightforward. It doesn't require injectivity of $f$. Here is a hint on proving the reverse inclusion: supposing $y \in f(S) \cap f(T)$, you'll have that $f(s) = y = f(t)$ for some $s \in S$ and $t \in T$, and since $f$ is injective, it must be that $s = t \in S \cap T$. – joeb Nov 07 '21 at 00:02

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