Compression function is not collision resistant but Merkle-Damgard is collision resistant Is it possible that you can still have a collision resistance in Merkle-Damgard even if the compression function has a collision?
Yes, a hash built per the Merkle-Damgård construction can be collision-resistant even if its  compression function has a known collision.

Consider SHA-256. Note its round function $F:\{0,1\}^{256}\times\{0,1\}^{512}\to\{0,1\}^{256}$ where the first argument is the state and the second is a message block. Now define $F'$ identical to $F$, except that $F'(0^{256},0^{512})$ is defined to be $F(0^{256},1^{512})$.

$F'$ has a known collision, yet the variant of SHA-256 using $F'$ is collision resistant, because we can't find a way to bring the state of SHA-256 to all-zero, which would essentially be a preimage attack.