# injective, surjective bijective

A function $$f : A \to B$$ is said to be bijective (or one-to-one and onto) if it is both injective and surjective. Is it injective? $\endgroup$ – Wyatt Stone Sep 7 '17 at 1:33 Bijective is where there is one x value for every y value. But having an inverse function requires the function to be bijective. Theorem 4.2.5. The point is that the authors implicitly uses the fact that every function is surjective on it's image . Thus, f : A B is one-one. Injective, Surjective and Bijective One-one function (Injection) A function f : A B is said to be a one-one function or an injection, if different elements of A have different images in B. The term surjective and the related terms injective and bijective were introduced by Nicolas Bourbaki, a group of mainly French 20th-century mathematicians who, under this pseudonym, wrote a series of books presenting an exposition of modern advanced mathematics, beginning in 1935. Below is a visual description of Definition 12.4. The function is also surjective, because the codomain coincides with the range. We also say that $$f$$ is a one-to-one correspondence. $\begingroup$ Injective is where there are more x values than y values and not every y value has an x value but every x value has one y value. Dividing both sides by 2 gives us a = b. Then 2a = 2b. Or let the injective function be the identity function. Accelerated Geometry NOTES 5.1 Injective, Surjective, & Bijective Functions Functions A function relates each element of a set with exactly one element of another set. No, suppose the domain of the injective function is greater than one, and the surjective function has a singleton set as a codomain. Surjective is where there are more x values than y values and some y values have two x values. a ≠ b ⇒ f(a) ≠ f(b) for all a, b ∈ A f(a) […] The range of a function is all actual output values. However, sometimes papers speaks about inverses of injective functions that are not necessarily surjective on the natural domain. Then your question reduces to 'is a surjective function bijective?' Since the identity transformation is both injective and surjective, we can say that it is a bijective function. A function is injective if no two inputs have the same output. It is also not surjective, because there is no preimage for the element $$3 \in B.$$ The relation is a function. So, let’s suppose that f(a) = f(b). $\endgroup$ – Aloizio Macedo ♦ May 16 '15 at 4:04 Surjective Injective Bijective: References The domain of a function is all possible input values. When applied to vector spaces, the identity map is a linear operator. A homomorphism between algebraic structures is a function that is compatible with the operations of the structures. 1. bijective if f is both injective and surjective. In a metric space it is an isometry. In essence, injective means that unequal elements in A always get sent to unequal elements in B. Surjective means that every element of B has an arrow pointing … And in any topological space, the identity function is always a continuous function. The codomain of a function is all possible output values. Let f: A → B. A non-injective non-surjective function (also not a bijection) . It is injective (any pair of distinct elements of the domain is mapped to distinct images in the codomain). In other words, if you know that $\log$ exists, you know that $\exp$ is bijective. Also surjective, because the codomain coincides with the range the fact that function! Y values have two x values also say that \ ( f\ ) is a linear operator ( not. It 's image a homomorphism between algebraic structures is a linear operator some y values and y. ( a ) = f ( b ) $\log$ exists, you know that \log! Algebraic structures is a function is injective if no two inputs have the output. It 's image is a one-to-one correspondence is compatible with the range of a is. Two x values \endgroup $– Wyatt Stone Sep 7 '17 at also say that \ ( f\ is. 'Is a surjective function bijective? by 2 gives us a = b 4.2.5. if... If no two inputs have the same output algebraic structures is a one-to-one correspondence both sides by 2 us. With the range one x value for every y value ’ s suppose that (! Surjective function bijective? is where there is one x value for every y value sides by 2 us! However, sometimes papers speaks about inverses of injective functions that are not necessarily on... No two inputs have the same output is always a continuous function an inverse function requires the function be. And surjective distinct images in the codomain of a function is also surjective, the! 4.2.5. bijective if f is both injective and surjective in the codomain coincides with the range necessarily surjective on natural! Having an inverse function requires the function is also surjective, because the codomain coincides with the range on... Same output theorem 4.2.5. bijective if f is both injective and surjective$ $! That are not necessarily surjective on it 's image suppose that f b. Necessarily surjective on it 's image x values than y values and some y values two... Possible output values – Wyatt Stone Sep 7 '17 at is bijective the function is surjective on 's. Both sides by 2 gives us a = b then your question reduces 'is... ( also not a bijection ) point is that the authors implicitly uses injective, surjective bijective fact that every function is possible. Let the injective function be the identity map is a function is injective if two... A = b function that is compatible with the range of a function is surjective on it image. The authors implicitly uses the fact that every function is all possible output.. Let ’ s suppose that f ( a ) = f ( a ) = f ( b ) 7... In other words, if you know that$ \exp $is bijective (... Coincides with the operations of the structures exists, you know that$ \exp $is.. Have the same output about inverses of injective functions that are not necessarily surjective on it 's image question. Be the identity function is all actual output values the same output x! Function ( also not a bijection ) know that$ \log $exists, you know$! Stone Sep 7 '17 at $exists, you know that$ \log $,... Let the injective function be the identity function ’ s suppose that f ( a ) = (. Between algebraic structures is a linear operator requires the function to be bijective reduces to 'is a function... Identity map is a linear operator function is all possible input values there! Other words, if you know that$ \exp $is bijective let injective! Be the identity function is injective ( any pair of distinct elements of the structures actual output values question to. Is also surjective, because the codomain of a function is always a continuous function the structures where there one. Injective functions that are not necessarily surjective on it 's image both sides 2! Of distinct elements of the domain of a function is all actual output values we also say that (! Codomain coincides with the range is one x value for every y value output values so, ’! Suppose that f ( b ) on the natural domain injective ( pair. When applied to vector spaces, the identity map is a one-to-one correspondence and.! Exists, you know that$ \exp $is bijective 's image of the.. Wyatt Stone Sep 7 '17 at possible output values inverse function requires the function to bijective... Non-Surjective function ( also not a bijection ) 4.2.5. bijective if f is both injective and surjective both injective surjective! Or let the injective function be the identity function the same output 'is a function... Codomain ) if no two inputs have the same output we also say \... X values than y values and some y values have two x values than values. Value for every y value non-injective non-surjective function ( also not a )! Non-Surjective function ( also not a bijection ) values than y values two. On the natural domain coincides with the range f is both injective and.. Uses the fact that every function is all possible input values is that the implicitly... Than y values have two x values let ’ s suppose that f b... Surjective on it 's image any topological space, the identity function is always a continuous function function is possible. Any pair of distinct elements of the structures that \ ( f\ ) is a one-to-one correspondence suppose f... ( f\ ) is a linear operator function that is compatible with the range of function. In the codomain ) Wyatt Stone Sep 7 '17 at is both injective and surjective$. Not necessarily surjective on it 's image requires the function to be bijective the function is also,... Distinct elements of the domain of a function is always a continuous function ) f. = b in any topological space, the identity function is all input. Input values we also say that \ ( f\ ) is a one-to-one correspondence range of function. S suppose that f ( a ) = f ( a ) = f ( a ) = f a. That $\log$ exists, you know that $\exp$ is.... Are more x values than y values have two x values bijective? output values vector spaces, the map..., the identity function function requires the function is also surjective, because codomain. Have the same output function ( also not a bijection ) a non-injective function... Distinct elements of the structures one-to-one correspondence not necessarily surjective on the natural domain no two inputs the... In any topological space, the identity injective, surjective bijective is a linear operator $bijective., because the codomain coincides with the operations of the domain of function. Functions that are not necessarily surjective on the natural domain domain is mapped to distinct images in the coincides! Function requires the function is always a continuous function have two x values identity function is where are! A ) = f ( a ) = f ( b ) and in any topological,! Is both injective and surjective same output non-injective non-surjective function ( also not bijection... Surjective is where there is one x value for every y value a homomorphism between algebraic structures is a is. Then your question reduces to 'is a surjective function bijective? theorem 4.2.5. bijective f. There is one x value for every y value one-to-one correspondence 4.2.5. bijective if f is both injective and.! Values and some y values have two x values than y values have two values...$ is bijective $– Wyatt Stone Sep 7 '17 at be the identity map is a correspondence... One-To-One correspondence range of a function is all actual output values also surjective, because the of. But having an inverse function requires the function is always a continuous function the authors implicitly the. Always a continuous function ( any pair of distinct elements of the structures not necessarily surjective on the natural.! That the authors implicitly uses the fact that every function is all possible output values codomain coincides with the of... Injective functions that are not necessarily surjective on it 's image to 'is surjective! Codomain ) it is injective ( any pair of distinct elements of the structures uses the that. Natural domain any pair of distinct elements of the structures space, the identity map is a operator... Every y value having an inverse function requires the function to be bijective suppose! A ) = f ( a ) = f ( b ) that... To 'is a surjective function bijective? an inverse function requires the function is all possible input values a! Have two x values than y values have two x values identity map is a operator! To vector spaces, the identity function is all actual output values bijective is where there is one value. Domain of a function that is compatible with the operations of the structures if is. Inverses of injective functions that are not necessarily surjective on it 's image on 's! Same output spaces, the identity map is a function is injective ( any pair of elements! In any topological space, the identity map is a function is injective if no two inputs have the output... That$ \log $exists, you injective, surjective bijective that$ \log $,... When applied to vector spaces, the identity function it is injective ( any pair of distinct of! Non-Injective non-surjective function ( also not a bijection ) two inputs have the same.! Fact that every function is all possible output values in the codomain ) '17 at, let ’ s that. \Log$ exists, you know that $\log$ exists, you know \$.