Is the Intermediate Value Theorem basically saying that if a function is continuous on an interval, then the function is surjective?The formal definition states something to the effect of "any value in the domain will map to a value in the range", unless I misunderstand it. So, since we're mapping from [ a , b ] → [ f ( a ) , f ( b ) ], the range is made up of the images of the values in the domain, that is, the range is the codomain.My understanding of a surjective function is one in which ∀ x ∈ D , f ( x ) ∈ C, where f : D → C.Or an I horribly mistaken?

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Is the Intermediate Value Theorem basically saying that if a function is continuous on an interval, then the function is surjective?The formal definition states something to the effect of &quot;any value in the domain will map to a value in the range&quot;, unless I misunderstand it. So, since we&#039;re mapping from   [  a  ,  b  ]  →  [  f  (  a  )  ,  f  (  b  )  ], the range is made up of the images of the values in the domain, that is, the range is the codomain.My understanding of a surjective function is one in which   ∀  x  ∈  D  ,  f  (  x  )  ∈  C, where   f  :  D  →  C.Or an I horribly mistaken?

Irene Simon · Accepted Answer

No. It means its image is an interval, but that&#039;s all: for instance, look at the constant function   f  :  [  0  ,  1  ]  →      R   defined by   f  (  x  )  =  1 for all   x  ∈  [  0  ,  1  ].It is continuous, and the IVT applies. But it is definitely not surjective... its image is the (trivial) interval   {  1  }. Surjectivity would ask that   ∀  y  ∈  C  =      R  , there exists   x  ∈  D  =  [  0  ,  1  ] such that   f  (  x  )  =  y.

metodystap9 · Accepted Answer

Let   I be the domain of   f. If we restrict the co-domain to the image   f  [  I  ], that is consider       f    ∗    :  I  →  f  [  I  ] given by       f    ∗    (  x  )  =  f  (  x  ), then       f    ∗   is surjective since a function trivially surjects onto its image.

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