Can we use a function not explicitly defined, for its properties alone? Side question to ($G \simeq H/ \lbrace e \rbrace \implies G \simeq H$)

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I was motivated to prove this in order to shore up my understanding of isomorphisms, but I came across what (I thought) is a more interesting question. Is it alright to use a function (an isomorphism $\pi$ in this case) that is not explicitly defined? I know it exists, and I know some other things about it, but it feels strange to use a function and not know exactly what it is. Is this "technique" ok? Anyway, here's what I originally came here to do, to check my understanding by writing this proof:

To show $G \simeq H$ we will construct a bijective homomorphism from $G$ to $H$.

Since $G \simeq H/ \lbrace e \rbrace$, there exists an isomorphism $\pi :G \rightarrow H/ \lbrace e \rbrace$. Let $\phi : G \rightarrow H$ be a function defined by $\phi(g_i) = h_{g_i}$, where $h_{g_i} \in \pi(g_i)$. To see $\phi$ is a homomorphism, let $g_m,g_n \in G$. Then $\phi(g_mg_n) = h_{g_{m}g_{n}} \;$, and since $\pi$ is a homomorphism, $h_{g_{m}g_{n}} = h_{g_m}h_{g_n}$, which shows $\phi(g_mg_n) = \phi(g_m) \phi(g_n)$.

To see $\phi$ is bijective, we first note that $\mid H\mid$ $=[\;H: \lbrace e \rbrace \;]$, and that $[\;H: \lbrace e \rbrace \;] =$ $\mid G \mid$ by our assumption. Now let $\phi(g_m) = \phi(g_n)$ with $g_m,g_n \in G$. Then, $h_{g_m} = h_{g_n}$ and this implies $g_m = g_n$ as $\pi$ is an isomorphism. Since $\mid H\mid$ $=$ $\mid G\mid$, and $\phi$ is onto, $\phi$ is bijective, and we are done.

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It is pretty common to use objects without knowing what they really are. It's actually so common that most of modern mathematics defines things not by what they are, but by what properties they have (see, for instance, the axioms of group theory, with the properties closure, associativity, identity and inverse, saying nothing about what a group really is).

Even in cases where you don't know something (say, a function with certain properties) exists, you can make arguments about what all such functions must do. If it leads to a contradiction, you've proven that none can exist (this is actually the main approach to showing that something can't exist). If you get something interesting, and later someone proves such functions do exist, then there is no problem, and whatever interesting thing you've found turned out to be usable.