I'm having an issue of (seemingly) conflicting information from textbooks involving mappings between topological spaces. If $ \pi : S \rightarrow S/\sim $ is the projection of a topology S into a quotient over the relation $ \sim $, the topology of $ S $ is transferred to the quotient by requiring that all sets $ V \in S / \sim \, $ are open if $ \pi^{-1} (V) $ are open in $ S $. I'm assuming that this mapping is probably defined such that $ \pi^{-1}(V) := \{ x \in S \, | \, \pi(x) = [x] \; \; \forall [x] \in V \subset S/\sim \}$.

In Tu's manifolds (page 72), this condition constitutes a topology because:

$$ \pi^{-1} \Big( \bigcup_\alpha V_\alpha \Big) = \bigcup_\alpha \pi^{-1} (V_\alpha) \\ \pi^{-1} \Big( \bigcap_\alpha V_\alpha \Big) = \bigcap_\alpha \pi^{-1} (V_\alpha) $$

for an indexed collection of sets $ V_\alpha \in S/\sim $. However, Baker's Introduction to Topology (page 22) states that for a mapping between sets $ f:X \rightarrow Y $ and a collection of indexed sets $ U_\alpha \in X $,

$$ f \Big( \bigcap_\alpha U_\alpha \Big) \subseteq \bigcap_\alpha f(U_\alpha ) $$

and that equality of these two sets is satisfied only when $ f $ is one-to-one. However, when this is applied to the mapping $ \pi^{-1} $ above, it is not only clear that $ \pi^{-1} $ isn't one-to-one, it isn't even well defined as a function, because it's very probable that many elements of $ S $ will map to a single equivalence class in $ S/\sim $, making the inverse multi-valued. Additionally, I can't seem to find any flaw in the following proof that the missing $ \subseteq $ relationship is true all the time:

$$ \text{Let } x \in \bigcap_\alpha \pi^{-1} (V_\alpha) \implies x \in \pi^{-1}(V_\alpha) \quad \forall \alpha \\ \implies \exists [x] \in V_\alpha \ni \pi(x) = [x] \quad \forall \alpha \implies [x] \in \bigcap_\alpha V_\alpha \\ \implies x \in \pi^{-1} \Big( \bigcap_\alpha V_\alpha \Big) \implies \bigcap_\alpha \pi^{-1} (V_\alpha) \subseteq \pi^{-1} \Big( \bigcap_\alpha V_\alpha \Big) $$

So is my proof valid, and Baker is simply wrong? I highly doubt that, it seems I'm missing something critical on why Tu is still correct, and Baker's theorem doesn't apply to this situation? I've seen a similar proof to the above posted here before, I guess my main question is why Baker's one-to-one restriction on the equality isn't killing my proof and as a result Tu's definition.


2 Answers 2


If we have a function $\phi : A \to B$ between two sets $A,B$, then we can define for each subset $M \subset A$ its image $$\phi(M) = \{ \phi(a) \mid a \in M \} \subset B$$ and for each subset $N \subset B$ its preimage $$\phi^{-1}(N) = \{ a \in A \mid \phi(a) \in N \} \subset A.$$ $\phi^{-1}$ is not a function from $B$ to $A$. It is actually a function $$\phi^{-1} : \mathfrak P (B) \to \mathfrak P (A), $$ where $\mathfrak P (X)$ denotes the powerset of the set $X$. Alternatively we could interpret $\phi^{-1}$ as "multivalued function from $B$ to $A$". Note that for a bijection $\phi : A \to B$ one usually writes $\phi^{-1} : B \to A$ for the inverse function. Thus we have the same notation for two different things:

  1. $\phi^{-1} : \mathfrak P (B) \to \mathfrak P (A)$ which is defined for all functions $\phi : A \to B$

  2. $\phi^{-1} : B \to A$ which is defined only for bijections $\phi : A \to B$

This may be confusing, but if you are used to it you will not regard it as a problem. The interpretation of $\phi^{-1}$ is normally clear from the context.

Similarly one writes $$\phi : \mathfrak P (A) \to \mathfrak P (B)$$ for the "image function" determined by $\phi$ which is also a little ambiguous.

Concerning unions the functions $\phi : \mathfrak P (A) \to \mathfrak P (B)$ and $\phi^{-1} : \mathfrak P (B) \to \mathfrak P (A)$ have the same behavior: $$\phi(\bigcup_\alpha M_\alpha) = \bigcup_\alpha \phi(M_\alpha) \phantom{x}, \phantom{x} \phi^{-1}(\bigcup_\alpha N_\alpha) = \bigcup_\alpha \phi^{-1}(N_\alpha) .$$ However, concerning intersections their behavior is different: $$\phi(\bigcap_\alpha M_\alpha) \subset \bigcap_\alpha \phi(M_\alpha) \text{ but in general } \phi(\bigcap_\alpha M_\alpha) \ne \bigcap_\alpha \phi(M_\alpha) \phantom{x}, \phantom{x} \phi^{-1}(\bigcap_\alpha N_\alpha) = \bigcap_\alpha \phi^{-1}(N_\alpha) .$$ Baker is right, the injectivity of $\phi : A \to B$ is a necessary and sufficent condition for $$(*) \phantom x \phi(\bigcap_\alpha M_\alpha) = \bigcap_\alpha \phi(M_\alpha). $$ To see that it is necessary, consider a non-injective function $\phi :A \to B$. There are $a, a' \in A$ such that $a \ne a'$ and $\phi(a) = \phi(a') = b$. Then $\emptyset = \phi(\emptyset) = \phi(\{a\} \cap \{a'\}) \ne \{b\} = \phi(\{a\}) \cap \phi(\{a'\})$. To see that it is sufficient, let $b \in \bigcap_\alpha \phi(M_\alpha)$, i.e. $b \in \phi(M_\alpha)$ for all $\alpha$. Thus $b = \phi(a_\alpha)$ for suitable $a_\alpha \in M_\alpha$. But since $\phi$ is injective we must have $a_\alpha = a_{\alpha'}$ for all $\alpha,\alpha'$. This means $a_\alpha = a$ for all $\alpha$. Clearly $a \in \bigcap_\alpha M_\alpha$, thus $b \in \phi(\bigcap_\alpha M_\alpha)$.

Your proof for $\phi^{-1}$ is correct and you do not need the injectivity of $\phi : A \to B$. But there is no contradiction: It only means that different functions (like $\phi : \mathfrak P (A) \to \mathfrak P (B)$ and $\phi^{-1} : \mathfrak P (B) \to \mathfrak P (A)$) may behave in a different manner.

  • $\begingroup$ I corrected my answer, the claim about constant functions was nonsense. $\endgroup$
    – Paul Frost
    Commented Dec 3, 2020 at 13:05
  • $\begingroup$ The distinction between definitions 1. and 2. clarify the situation for me greatly, thank you. The concept of pre-image is now perfectly clear, I'm just embarrassed that I somehow have avoided true understanding of the concept until now. $\endgroup$
    – Jerome
    Commented Dec 3, 2020 at 20:38

Preimages of maps behave differently, in some sense "nicer" if you want, than images. One of my professors always used to say

preimages are good, images are evil.

What he meant is that, given a map $f\colon X\to Y$ with a collection of subsets $A_i \subset X, B_i \subset Y$ for an index set $I$, preimages respect unions and intersections $$f^{-1}\left(\bigcap_i B_i\right) = \bigcap_if^{-1}(B_i)\quad \text{and}\quad f^{-1}\left(\bigcup_i B_i\right) = \bigcup_if^{-1}(B_i)$$ the images, in general, do not:

$$f\left(\bigcap_i A_i\right) \subseteq \bigcap_i f(A_i)$$

And, as you already figured, they are equal iff $f$ is injective.

However, you still seem to be confusing a few things. The projection $\pi$ is for obvious reasons not injective but it's surjective and it's certainly well defined. You are confusing preimages $$\pi^{-1}([s])\subset S$$ of elements $[s]\in S/{\sim}$ under $\pi$, which are subsets of $S$, with the inverse function $$\pi^{-1}\colon S/{\sim}\to S.$$ These are entirely different things.

The projection $\pi$ projects all elements $a,a' \in S$ that are equal $a\sim a'$ with respect to the equivalence relation $\sim$ onto their common equivalence class $[a]$ in $S/{\sim}$.

  • $\begingroup$ My apologies, I meant to imply that $ \pi^{-1} $ isn't well-defined as a single-valued function, not $ \pi $. I think I see now, though, thank you! $\endgroup$
    – Jerome
    Commented Dec 3, 2020 at 20:34

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