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March 20

Whether it coincides with a simpler function

Is Template:Math (1) the same function as y=x, if we consider all branches of logarithm (of any real number) and all branches of inverse sine function? Or does (1) remain meaningless for any argument outside the range [-1;1] when we restrict it to real value for both the domain and the image, and (1) will coincide with the identity function only when we regard it as a function that map complex numbers to complex number? Does the logarithm of negative numbers lead to the presence of removable singularities for (1)? (In contrast, the function y=x obviously does not contain any singularity). I was able to prove that Template:Math and Template:Math are not always the same, but I still can't settle the aforementioned problems. 2402:800:63AD:81DB:105D:F4F:3B26:74C5 (talk) 14:36, 20 March 2024 (UTC)

A univalued function and a multivalued function $f:X\to Y,$ possibly partial, can be represented by a relation $R_f\subseteq X\times Y.$ The total identity function $\text{id}:X\to X$ corresponds to the identity relation $I_X=\{(x,x)|x\in X\}\subseteq X\times X.$ Function composition corresponds to relation composition: $R_{(g\circ f)}=R_f;R_g.$ The multivalued function inverse correspond to relation converse: $R_{(f^{-1})}=(R_f{)}^{\text{T}}.$

Just like the multivalued complex logarithm is the multivalued inverse of the exponential function $\exp :ℂ\to ℂ$, the complex $\arcsin$ including all branches is the multivalued inverse of function $\sin :ℂ\to ℂ.$ So

$R_{(\sin \circ \arcsin )}=R_{\arcsin };R_{\sin }=(R_{\sin }{)}^{\text{T}};R_{\sin }.$

Generalizing this from the sine function to an arbitrary (univalued) function $f$, we have:

$(x,y)\in (R_f{)}^{\text{T}};R_f\iff \mathrm{\exists }z:((z,x)\in R_f)\wedge (z,y)\in R_f)\iff \mathrm{\exists }z:(x=f(z)\wedge y=f(z)).$

Clearly, this implies $x=y,$ so the composed relation is the identity relation on the range of $f,$ representing the identity function on that range.  --Lambiam 18:21, 20 March 2024 (UTC)