## Complexity of finding a Eulerian path such that the image under a bijection is also a Eulerian path

Problem input: undirected graphs $$G$$, $$H$$ and a bijection $$f: E(G) \to E(H)$$
Question: Is there a Eulerian path $$p: \{1,\dots,|E(G)|\} \to E(G)$$ in $$G$$ such that $$f \circ p$$ is a Eulerian path in $$H$$?

Is this problem in $$P$$, $$NP$$-complete or maybe equivalent to some other well-known problem that isn’t known to be either yet?

The problem turned up as a special case of a more general problem first in the Bachelors thesis of another student last year and now in mine as well (I was improving some of the results of the other thesis). All other cases of the problem are now known to be either NP-hard or solvable in polynomial time, but neither one of us managed to make any progress on that case. On the other hand it feels like a much more "natural" problem than the other cases, so I think we may have missed a simple proof.

If undirected graphs are replaced by directed graphs in the problem statement there is a simple polynomial time algorithm (find a Eulerian path in the graph with vertex set $$V(G)\times V(H)$$ and edge set $$\{((v_1, w_1), (v_2, w_2)) | (v_1, v_2)=e \in E(G), f(e) = (w_1, w_2)\}$$ ignoring isolated vertices). The analogous graph for the undirected case contains 2 edges for each edge in $$G$$, so just finding a Eulerian path is no longer enough.

## Not onto, not one-to-one, and not bijection

If there is a mapping of x to y, where more than one value of x can map to the same, and also not all the values of y have a mapping from x, is there a name for this sort of function? (note if not implied: x has less elements than y)

## Bijection between metric spaces

Let de be the usual Euclidean metric, dD be the discrete metric and R be the reals. Define a function f from (R,dD) to (R,de) by f(x) = x. Is f a bijection? Why? Is f continuous, uniformly continuous, a topological isomorphism and an isometry?

Note: (R,dD) and (R,de) are metric spaces.

## JProve $T_1$ is finite if there exists bijection from $T_1$ to another finite set $T_2$

Assuming $$T_1$$ is non empty. Prove $$T_1$$ is finite if there exists bijection from $$T_1$$ to another finite set $$T_2$$

Now we have $$h : J_m \to T_1$$

Assume another finite set $$T_2$$, so we have $$h_1 : J_m \to T_2$$.

Since both $$h_1$$ and $$h_2$$ are bijections. so is $$h_1 \circ h^{-1}$$

$$J_m$$ = $${1,2,3,…,m$$}

Conversely, similar idea to that of this. But is this even correct ?

Thanks

## is there a bijection for f: $\mathbb R$ $->$ $\mathbb C$?

I imagine no since the dimensions do not match

but they have the same cardinality $$|\mathbb R |= |\mathbb C|$$?

## Bijection $f: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$

I need a bijection, such that: $$f(x,y) = \phi(x) + \psi(y)$$ and $$f(x,y) = g(\phi(x) + \psi(y))$$. Second one is easy, I think. If we make first one, we can consider $$g$$ as identical map. I have seen similar topics on stack, but I don’t like those answers. I think it is possible to construct a bijection, as Cantor did, proving that $$[0,1] \times [0,1]$$ ~ $$[0,1]$$, because I can build a bijection $$h: [0,1] \to \mathbb{R}$$. Thus we have a equivalent problem, with segment, i.e build a bijection $$f: [0,1] \times [0,1] \to [0,1]$$, such that $$f(x,y) = \phi(x) + \psi(y)$$.