2022-11-13

391: Composition of Preimage After Map of Subset Is Identical If Map Is Injective with Respect to Argument Set Image

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that composition of preimage after map of subset is identical if map is injective with respect to argument set image

Topics


About: set

The table of contents of this article


Starting Context



Target Context


  • The reader will have a description and a proof of the proposition that for any map, the composition of the preimage after the map of any subset is identical if the map is injective with respect to the argument set image.

Orientation


There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.


Main Body


1: Description


For any sets, S1,S2, any map, f:S1S2, and any subset, S3S1, f1f(S3)=S3 if f is injective with respect to f(S3), which means that for any pf(S3), f1(p) is a 1 element set.


2: Proof


Suppose that f is injective with respect to f(S3). For any pf1f(S3), f(p)f(S3) by the definition of preimage, which means that there is a pS3 such that f(p)=f(p), but as f1(f(p)) is a 1 element set, p=p, so, pS3. For any pS3, f(p)f(S3), which means that pf1f(S3) by the definition of preimage.


3: Note


It is important not to carelessly conclude that f1f(S3)=S3 without checking the condition.

The condition is not any necessary condition; in fact, the issue is not really being injective, but f1f(S3)S3, which is a necessary condition; being injective is just a typical case of guaranteeing f1f(S3)S3.

The condition is 'with respect to f(S3)', not "with respect to S3": the condition is not about the injectivity of f|S3:S3S2.


References


<The previous article in this series | The table of contents of this series | The next article in this series>