# Uniqueness of the inverse kernel of an invertible integral transform

Say we have an integral transform $$T$$ that maps any function $$f$$ to the function $$\varphi=T\left\lbrace f\right\rbrace$$, defined by

$$\varphi(s)=\left[T\left\lbrace f\right\rbrace\right](s)=\int_a^bK(x,s)f(x)dx$$

With $$K$$ a given function of two variables, called the kernel of the transform.

The domain $$A$$ on which $$T$$ is defined is the set of all functions $$f$$ such that the integral is exists.

Suppose $$T$$ is injective, then it is bijective from $$A$$ to its range $$B=T\left( A\right)$$, and the inverse transform $$T^{-1}:B\to A$$, that recovers $$f$$ from $$\varphi$$, exists.

Now, for some integral transforms, I know that there is an analytical expression for $$T^{-1}$$ if $$\varphi$$ belongs to a certain subset of $$B$$:

$$f(x)=\left[T^{-1}\left\lbrace \varphi\right\rbrace\right](x)=\int_c^dK^{-1}(x,s)\varphi(s)ds$$

In which case, the function of two variables $$K^{-1}$$ is called the inverse kernel of $$K$$ (hence the $$^{-1}$$ exponent in the notation).

My question is, for the generic injective integral transform $$T$$ and its associated Kernel $$K$$ defined above, how can we tell that there is at most one possible inverse Kernel ? In other words, how can we tell that if there is an inverse kernel $$K^{-1}$$, it is necessarily unique ?

Naively, I can imagine a scenario where, although $$T^{-1}$$ is known to exist because of the injectivity of $$T$$, we don't have a analytical expression of $$T^{-1}$$ valid on all of its domain (i.e the range of $$T$$), but rather two integral transforms with each a different Kernel, defined on two disjoint subset of the domain $$B$$ of $$T^{-1}$$, such that $$T^{-1}$$ coincides with each of these on their respective domains...

Any insight ?

(if you think my question is ill-defined, feel free to suggest a more appropriate setting)