Linked Questions

0
votes
0answers
124 views

Suppose that f is a function from A to B, where A and B are finite sets |A| = |B|. Show that f is one-to-one if and only if it is onto. [duplicate]

Suppose that f is a function from A to B, where A and B are finite sets |A| = |B|. Show that f is one-to-one if and only if it is onto. Lets say we were to translate this statement using variables. ...
0
votes
0answers
49 views

Proving a function is bijection [duplicate]

My problem is: Let $A$ and $B$ finite sets of equal cardinality. Show that if a function $f : A \to B$ is injective then it is also surjective. Similarly show that if $f$ is surjective, then it is ...
0
votes
0answers
16 views

Cardinalities of the two sets and the condition of the function in between [duplicate]

So if I have two sets A and B and if |A| = |B| and if function f is a function from A to B and is injective, how can I prove that it is also surjective? Forgot to mention. A and B are finite sets.
12
votes
2answers
16k views

Surjectivity implies injectivity

Let S be a finite set.Let F be a surjective function from S to S. How do I prove that it is injective?
4
votes
3answers
2k views

Empty functions are not injective?

Many sources say that empty functions such as $f:\emptyset \rightarrow S$ are injective because it is a vacuous truth. But currently I am reading a book on axiomatic set by Patrick Suppes, and he ...
1
vote
1answer
761 views

Counting the number of injective and surjective functions from a set to itself

I have a question about counting the number of injective (one-to-one) and surjective (onto) functions from $\{1,2,...,n\}$ to itself that satisfy $|f(i)-i| \leq 1, \forall i\in \{1,2,...,n\}$. I ...
0
votes
3answers
610 views

Showing that if $f$ is surjective, then $m\geq n$ holds (where $m$ and $n$ are the number of elements in the domain and codomain respectively)

Let $X$ and $Y$ be sets with $m$ and $n$ elements respectively and let $f:X\rightarrow Y$ be a function from $X$ to $Y$. Show that if $f$ is surjective, then $m\geq n$. I know this should be ...
0
votes
2answers
179 views

Non-bijective endomorphisms of finite groups

How can I prove that there are no finite groups $G$ where there exists an endomorphism $G\rightarrow G$ that is injective but not surjective, or other way round?
2
votes
2answers
139 views

Prove a map X → X must be bijective

Let X be a finite set. Does there exist a map $α : X → X$ such that $α$ is surjective, but not injective? $$ \text{No.} $$ $$\begin{align} \text{Surjective: }& \ \ {\displaystyle \forall y\in Y,\...
1
vote
3answers
135 views

How to write a rigorous proof for this statement?

Prove that for finite set $X$, the function $f:X \to X$ is surjective if and only if it is injective I have the idea of proof in my mind but find it difficult to translate it into mathematical ...
1
vote
2answers
61 views

is it true that an injective unary operation on a finite set is surjective?

To prove the statement:Any finite integral domain $R$ is a field. Let $r \in R$ , we define a function$ f: R \rightarrow R$ by $f(x)=rx$. The proof says: this map is injective(i understand this), then ...
1
vote
3answers
95 views

Evaluating correctness of various definitions of countable sets

I was trying to understand the definition of countable set (again!!!). Wikipedia has a very great explanation: A set $S$ is countable if there exists an $\color{red}{\text{injective}}$ function $...
1
vote
1answer
114 views

If mapping is surjective, then it's injective in finite sets.

If the set has a finite number of elements, prove the following: If $\sigma$ maps $S$ onto $S$, then $\sigma$ is one-to-one. Proof: Suppose $\sigma$ is not 1-1, then $\exists x_1, x_2$ $(x_1\neq x_2)...
0
votes
1answer
63 views

Definition of terms for functions

Hi I have this question from my homework and I'm stuck on it. It says: Let X and Y be sets. Suppose that X is finite and let f : X → Y be a map. Show that f is injective if and only if |f(X)| = |X|. ...
1
vote
0answers
45 views

Prove, if 𝑓 is surjective function, then 𝑓 is injective function.

The question would be: Let 𝑈 – finite set, and 𝑓:𝑈 → 𝑈. Prove, if 𝑓 is a surjective function, then 𝑓 is an injective function. Proposed solution: Consider f is not an injective function. ...