Claim. If $G$ is a group with a normal subgroup $K$, then $G/K \times K \cong G$.
Proof. $G/K$ is quotient group consisting of equivalence classes. Pick a representative $[g]$ for each equivalence class. Then consider the map $\varphi : G / K \times K \longrightarrow G$ defined by $([g], k) \mapsto gk$. The map is well-defined since we picked a single representative for each equivalence class. If $\varphi([g], k) = gk = g'k' = \varphi([g'], k')$, then $g = g' k' k^{-1} \in g'K$, so $[g] = gK = g'K = [g']$ and thus $g = g'$ since we have chosen one specific representative. It follows from $g = g'$ and $gk = g'k'$ that $k = k'$, so $([g], k) = [g'], k')$; therefore $\varphi$ is injective. Clearly $\varphi$ is surjective: if $x \in G$, then $x \in [g]$ for some representative $g \in G$, so $x = g\tilde{k}$ for some $\tilde{k} \in K$, so $\varphi([g], \tilde{k}) = g \tilde{k} = x$. $\tag*{$\square$}$ The proof is fake and the claim is false. All we have is a bijection between $G/K \times K$ and $G$. There are counterexamples: Is $ G \cong G/N \times N$?. Also relevant: When does the isomorphism $G\simeq ker(\phi)\times im(\phi)$? hold?.
Suppose $V, W$ are vector spaces over a field $K$. Let $\phi : V \to W$ be a linear transformation. If $V$ is finite-dimensional, then $V \cong \ker \phi \times \text{im} \, \phi$ via some isomorphism $\tau$. Then we can decompose $\phi$ as $\phi = i \circ p \circ \tau$, where $p: \ker \phi \times \text{im} \, \phi \to \text{im} \, \phi$ is projection $(a, b) \mapsto b$ and $i : \text{im} \, \phi \to W$ is inclusion $w \mapsto w$. So when $V$ is finite-dimensional, any linear transformation $\phi : V \to W$ is a composition of isomorphisms, inclusions, and projections. I want to prove that this fact is still true when $V$ is infinite-dimensional. It suffices to prove that if $\phi : V \to W$ is linear, then $V \cong \ker \phi \times \text{im} \, \phi$.
My first thought was to use $\text{im} \, \phi \cong V / \ker \phi$ and conclude $\text{im} \, \phi \times \ker \phi \cong V / \ker \phi \times \ker \, \phi \cong V$. But based on the above fake fact, I don't immediately expect the last equality to be true. It turns out that if $V/K$ is finite-dimensional, then $V/K \times K \cong V$ (see here). So I don't know how to handle the case that both $V$ and $V/\ker \phi$ are infinite-dimensional.
I thought I would find an answer at Is a vector space isomorphic to the kernel $\oplus$ image of a map out of it?. But that question seems to assume that the bijection in the fake proof above is a vector space isomorphism. Just before posting the question, I found an answer in the "similar questions" suggestion: if $\phi : V \to W$ is a linear transformation between vector spaces, then $V \cong \ker \phi \times \text{im} \, \phi$. But I am still curious if the original proof idea will work. Specifically:
If $V$ is an infinite-dimensional vector space and $K$ is a subspace of $V$ such that $V/K$ is infinite-dimensional, then is $V/K \times K \cong V$?
